Selecting an Indian food item from the given 10 can be done in 10 ways. Selecting a Chinese food item from the given 7 can be done in 7 ways.
∴ Selecting an Indian or Chinese food can be done in 10 + 7 = 17 ways.
(ii) There are 3 types of a toy car and 2 types of toy train are available in a shop. Find the number of ways a baby can buy a toy car and a toy train?
Number of types of Toy car = 3
Number of types of Toy Train = 2
Number of ways of buying a Toy car = 3 ways
Number of ways of buying a toy train = 2 ways
∴ By fundamental principle of multiplication, number of ways of buying a toy car and a toy train = 3 × 2 ways = 6 ways
(iii) How many two – digit numbers can be formed using 1, 2,3,4,5 without repetition of digits?
The given digits are 1, 2, 3, 4, 5
A two-digit number has a unit place and 10’s place. We are given 5 digits (1, 2, 3, 4, 5). The unit place can be filled (using the 5 digits) in 5 ways. After filling the unit place since repetition is not allowed one number (filled in the unit place) should be excluded. So the 10’s place can be filled (using the remaining 4 digits) in 4 ways.
∴ Unit place and 10’s place together can be filled in 5 × 4 = 20 ways. So the number of two-digit numbers = 20
(iv) Three persons enter into a conference hall in which there are 10 seats. In how many ways can they take their seats?
Number of seats in the conference hall = 10
Number of persons entering into the conference hall = 3
Number of ways of getting a seat for 1st person = 10
Number of ways of getting a seat for 2nd person = 9
Number of ways of getting a seat for 3rd person = 8
By fundamental principle of multiplication, number of ways of getting seats for 3 persons in conference hall = 10 × 9 × 8 ways = 720 ways
(v) In how ways 5 persons can be seated In a row?
Number of persons = 5
5 persons can be arranged among themselves in 5! ways
(i.e) 5 × 4 × 3 × 2 × 1 = 120 ways
Number of distinct digit in a passcode of a mobile phone = 6
First digit can be tried in 10 ways using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Second digit can be tried in 9 ways
Third digit can be tried in 8 ways
Fourth digit can be tried in 7 ways
Fifth digit can be tried in 6 ways
Sixth digit can be tried in 5 ways
Therefore, the maximum number of attempts made to retrieve the passcode = 10 × 9 × 8 × 7 × 6 × 5 = 151200
(ii) Given four flags of different colours, how many different signals can be generated if each signal requires the use of three flags, one below the other?
Number of flags = 4
Number of flags required for a signal = 3
The total number of signals is equal to the number of ways of filling 3 places in succession by 4 flags of different colours. Number of ways of filling the top place using 4 different colour flags is 4 ways. Number of ways of filling the middle place using the remaining 3 different colour flags is 3 ways. Number of ways of filling the bottom place using the remaining 2 different colour flags is 2 ways.
Therefore, by fundamental principle of multiplication, the total number of signals = 4 × 3 × 2 = 24 ways
(i) Number of children in the running race = 4
The first place can be filled in (from the 4 children) 4 ways
After filling in the first place only 3 children are left out
So the second place can be filled in (from the remaining 3 children) 3 ways
So the first and the second places together can be filled in 4 × 3 = 12 ways
(ii) The first and second places can be filled in 12 ways
The third-place can be filled (from the remaining 2 children) in 2 ways and the fourth place can be filled in 1 way
So the race can be finished in 12 × 2 × 1 = 24 ways
(i) Repetitions of digits is allowed
The given digits are 2, 4, 6, 8
A number of ways of filling the unit place using the 4 digits 2, 4, 6, 8 is 4 ways. Number of ways of filling the tens place using the 4 digits 2, 4, 6, 8 in 4 ways Number of ways of filling the hundred’s place using the 4 digits 2, 4, 6, 8 is 4 ways
Therefore, by fundamental principle of multiplication, the total number of 3 digit numbers = 4 × 4 × 4
= 64 ways
(ii) The unit place can be filled (using the 4 digits) in 4 ways after filling the unit place since repetition of digits is not allowed that digit should be excluded.
So the 10’s place can be filled in (4 – 1) 3 ways and the 100’s place can be filled in (3 – 1) 2 ways
So the unit place, 10’s and 100’s places together can be filled in 4 × 3 × 2 = 24 ways
(i.e) The number of 3 digit numbers = 4 × 3 × 2 = 24 ways
(i) With repetition:
The given digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
The unit place can be filled in only one way using the digit 3. The ten’s place can be filled in 10 ways using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. The hundred’s place can be filled in 9 ways using the digits 1, 2, 3, 4, 5, 6, 7, 8, 9.
Therefore, by the fundamental principle of multiplication, the total number of 3 digit numbers is = 1 × 10 × 9 = 90
(ii) Without repetition:
The digits are 0, 1,2, 3, 4, 5, 6, 7, 8, 9
A three-digit number has 3 digits l’s, 10’s, and 100’s place.
The unit place is (filled by 3) filled in one way.
After filling the unit place since the digit ‘0’ is there, we have to fill the 100’s place. Now to fill the 100’s place we have 8 digits (excluding 0 and 3) So 100’s place can be filled in 8 ways.
Now to fill the 10’s place we have again 8 digits (excluding 3 and any one of the number) So 10’s place can be filled in 8 ways.
∴ Number of 3 digit numbers with ‘3’ in unit place = 8 × 8 × 1 = 64
(i) Repetition of digit is allowed:
The numbers between 100 and 500 will have 3 digits. The unit place can be filled in 6 ways using the digits 0, 1, 2, 3, 4, 5. The ten’s place can be filled in 6 ways using the digits 0, 1, 2, 3, 4, 5. The hundred’s place can be filled in 4 ways using the digits 0, 1, 2, 3, 4, 5 ( Excluding 0 and 5). Therefore, by fundamental principle of multiplication, the number of 3 digit, numbers between 100 and 500 with repetition of digits using the digits 0, 1, 2, 3, 4, 5 is = 6 × 6 × 4 = 144
(ii) Repetition of digits is not allowed:
The 100’s place can be filled (by using 1, 2, 3, 4) 10’s in 4 ways
The 10’s place can be filled in (6 – 1) 5 ways and the unit place can be filled in (5 – 1) 4 ways
So the number of 3 digit number 4 × 5 × 4 = 80
The given digits are 0, 1, 2, 3, 4, 5
To find the possible 3 – digit odd numbers.
(i) Repetition of digits is not allowed:
Since we need 3 – digit odd numbers the unit place can be filled in 3 ways using the digits 1, 3 or 5. Hundred’s place can be filled in 4 ways using the digits 0, 1, 2, 3, 4, 5 excluding 0 and the number placed in unit place. Ten’s place can be filled in 4 ways using the digits 0, 1, 2, 3, 4, 5 excluding the digit placed in the hundred’s place.
Therefore, by the fundamental principle of multiplication, the number of 3 – digit odd numbers formed without repetition of digits using the digits 0, 1, 2, 3, 4, 5 is
= 4 × 4 × 3 = 48
(ii) Repetition of digits is allowed:
The unit place can be filled in 3 ways. We are given 6 digits.
So 10’s place can be filled in 6 ways and the 100’s place can be filled in (6 – 1) (excluding zero) 5 ways
So the Number of 3 digit numbers = 3 × 6 × 5 = 90
To find the numbers between 999 and 10,000 using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
To find the possible 4 digit numbers.
(i) No restriction:
Thousand’s place can be filled in 9 ways using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 excluding 0. Since there is no restriction, the hundred’s place, Ten’s place, and the unit place can be filled in 10 ways using the digit 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
Therefore, by the fundamental principle of multiplication, the number of 4 digit numbers between 999 and 10,000 is = 9 × 10 × 10 × 10 = 9000
(ii) No digit is repeated:
Thousand’s place can be filled in 9 ways using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 excluding 0. Since repetition is not allowed. The unit place can be filled in 9 ways using the digits 0,1, 2, 3, 4, 5, 6, 7, 8, 9 excluding the digit placed in the thousand’s place. Since repetition of digits is not allowed, the ten’s place can be filled in 8 ways using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 excluding the digits used in thousand’s a place and unit place. Since repetition of digits is not allowed, the hundred’s place can be filled in 7 ways using the digits 0,1,2,3,4, 5,6,7,8,9 excluding the digits used in thousand’s a place and unit place.
Therefore, by the fundamental principle of multiplication, the number of numbers between 999 and 10,000 without repetition of digits is = 9 × 7 × 8 × 9 = 4536
(iii) At least one of the digits is repeated:
Required number of 4 digit numbers = Total number of 4 digit numbers – Number of 4 digit numbers when no digit is repeated = 9000 – 4536 = 4464
The given digits are 0, 1, 2, 3, 4, 5
To find the 3 – digit numbers formed by using the digits 0, 1, 2, 3, 4, 5 which are divisible by 5.
(i)The repetition of digits are not allowed:
Since the 3 – digit number is divisible by 5, the unit place can be filled in 2 ways using the digits 0 or 5
Case (i) When the unit place is filled with the digit 0. The hundreds place can be filled in 5 ways using the digits 1, 2, 3, 4, 5 and the ten’s place can be filled in 4 ways using the digits 0, 1, 2, 3, 4, 5 excluding 0 and the digit which is placed in hundred’s place.
Therefore, by fundamental principle of multiplication, the number of 3 – digit numbers divisible by 5 is
= 5 × 4 × 1 = 20
Case (ii) When the unit place is tilled with the digit 5, since repetition of digit is not allowed the hundred’s place can be filled in 4 ways using the digits 1, 2, 3, 4 (0 and 5 are excluded). The ten’s place can be filled in 4 ways using the digits 0, 1, 2, 3, 4, 5 (excluding 5 and the digit placed in the hundred’s place).
Therefore, by the fundamental principle of multiplication, the number of 3 – digit numbers, in this case, is = 4 × 4 × 1 = 16
Therefore, the total number of 3 – digit numbers divisible by 5 using the digits 0, 1, 2, 3, 4, 5 is = 20 + 16 = 36
(ii) The repetition of digits are allowed:
The digits are
0 1 2 3 4 5
To get a number divisible by 5 we should have the unit place as 5 or 0 So the unit place (using 0 or 5) can be filled in 2 ways.
The 10’s place can be filled (Using 0, 1, 2, 3, 4, 5) in 6 ways and the 100’s place (Using 1, 2, 3, 4, 5) can be filled in 5 ways.
So the number of 3 digit numbers ÷ by 5 (with repetition) = 2 × 6 × 5 = 60
Route map diagram for the given data.
The possible choices for a number of routes commuting from A to place C via place B without using similar mode transportation are
(B 1, T’ 1 ), (B 1, T’ 2 ), ( B 1, A 1 ), ( B 2, T’ 1 ), (B 2, T’ 2 )
(B 2, A’ 1 ), (T 1, B’ 1 ), (T 1, A’ 1 ), ( T 2, B’ 1 ), ( T 2, A’ 1 ) (A 1, B’ 1 ), (A 1, T’ 1 ) and (A 1, T’ 2 )
Therefore, the Required number of routes is 13.
From 1 to 1000, the numbers ÷ by 2 = 500
the number ÷ by 5 = 200
and the numbers ÷ by 10 = 100(5 × 2 = 10)
So number ÷ by 2 or 5 = 500 + 200 – 100 = 600
Total numbers from 1 to 1000 = 1000
So the number of numbers which are ÷ neither by 2 nor by 5 = 1000 – 600 = 400
Three-digit numbers:
The unit place can be filed in 4 ways using the digits 1, 3, 7, 9. Hundred’s place can be filled in 9 ways excluding 0. Ten’s place can be filled in 10 ways using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Therefore, the required number of 3 digit numbers neither divisible by 2 nor by 5 is = 9 × 10 × 4 = 360.
There is only one 4 – digit number, but it is divisible by 2 and 5.
Therefore, required numbers using fundamental principle of addition = 4 + 36 + 360 = 400
(i) Either starts with L or ends with S
The first box is filled with the letter L. The second box can be filled with the remaining letters O, T, U, S in 4 ways. The third box can be filled with the remaining letters excluding L and the letter placed in box 2 in 3 ways. The fourth box can be filled with the remaining letters excluding L and the letters placed in a box – 2 and box – 3 in 2 ways. The fifth box can be filled with the remaining one letter excluding L and the letters placed in a box – 2 and box – 3, box – 4 in 1 way.
Therefore, by fundamental principle of multiplication, the number of words start with L is = 1 × 4 × 3 × 2 × 1 = 24
Since the word ends with S, the fifth box can be filled in one way with the letter S. The remaining four boxes can be filled 4 × 3 × 2 × 1 way.
Therefore, the number of words ending with S = 4 × 3 × 2 × 1 × 1 = 24
Number of words starting with L and ends with S: The first box can be filled with L in one way fifth box can be filled with S in one way second box, third box, and fourth box can be filled in 3x2x1 ways with the remaining letters O, T, U.
∴ Number of words starting with L and ends with S = 1 × 3 × 2 × 1 × 1 = 6
Therefore, by fundamental principle of addition, number of words either starts with L or ends with S = 24 + 24 – 6 = 48 – 6 = 42
(ii) Neither starts with L nor ends with S Total number of words formed using the letters L, O, T, U, S is = 5 × 4 × 3 × 2 × 1 = 120
The number of words neither starts with L nor ends with S = Total number of words – Number of words starts with either L or ends with S = 120 – 42 = 78
MCQ Questions For Class 11 hindi with answers·
Count the total number of ways of answering 6 objective type questions, each question having 4 choices.
One question can be answered in 4 ways
Two questions can be answered in 4 × 4 = 4 2 ways
∴ Six questions can be answered in 46 ways
(ii) In how many ways 10 pigeons can be placed in 3 different pigeon holes?
First pigeons can be placed in a pigeon-hole in 3 ways (selecting 1 from 3 holes)
Second pigeons can be placed in a pigeon-hole in 3 ways Tenth pigeons can be placed in a pigeon-hole in 3 ways
So total number of ways in which all the number 10 place can be sent = 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 = 310 ways
(iii) Find the number of ways of distributing 12 distinct prizes to 10 students?
To give the first prize we have to select, from the 10 students which can be done in 10 ways.
To give the second prize we have to select one from the 10 students which can be done is 10 ways.
To give the 12th prize we have to select one from 10 students which can be done in 10 ways.
So all the 12 prizes can be given in (10 × 10 × 10 …. 12 times) = 1012 ways.
(i) 6!
6! = 6 × 5 × 4 × 3 × 2 × 1 = 720
(ii) 4! +5!
4! +5! = (4 × 3 × 2 × 1) + (5 × 4 × 3 × 2 × 1)
= (4 × 3 × 2 × 1) [1 + 5]
= 24 × 6 = 144
(iii) 3! – 2!
3! – 2! = (3 × 2 × 1) × (4 × 3 × 2 × 1)
= 6 × 24 = 144
(iv) 3! × 2!
3! × 2! = (3 × 2 × 1) × (4 × 3 × 2 × 1)
= 6 × 24 = 144
(v) \(\frac{12 !}{9 ! \times 3 !}\)
= 2 × 11 × 10 = 220
(vi) \(\frac{(n+3) !}{(n+1) !}\)
= (n + 3) (n + 2)
= n 2 + 3n + 2n + 6
= n 2 + 5n + 6
(i) n = 6, r = 2
(ii) n = 10, r = 3
(iii) For any n with r = 2.
(i) ( n + 1) ! = 20 ( n – 1 )!
(n + 1) n(n – 1)! = 20(n – 1)!
n(n + 1) = 20
n 2 + n – 20 = 0
n 2 + 5n – 4n – 20 = 0
n(n + 5) – 4(n + 5) = 0
(n – 4) (n + 5) = 0
n – 4 = 0 or n + 5 = 0
n = 4 or n = -5
But n = -5 is not possible. ∴ n = 4
(ii) \(\frac{1}{8 !}+\frac{1}{9 !}=\frac{n}{10 !}\)
Given (n – 1)P 3: nP 4 = 1: 10
Given 10P r – 1 = 2 × 6P r
5 × 9 × 8 × 7 = (11 – r) (10 – r) (9 – r) (8 – r) (7 – r)
5 × 3 × 3 × 2 × 4 × 7 = (11 – r) (10 – r) (9 – r) (8 – r)(7 – r)
7 × 6 × 5 × 4 × 3 = ( 11 – r ) (10 – r) (9 – r) (8 – r) (7 – r)
(11 – 4) (10 – 4) (9 – 4) (8 – 4) (7 – 4)
= (11 – r) (10 – r) (9 – r) (8 – r) (7 – r)
∴ r = 4
From 8 persons we have to select and arrange 3 which can be done in 8 P 3 ways So the prizes can be awarded in 8 P 3 = 8 × 7 × 6 = 336 ways
(ii) Three men have 4 mats, 5 waist coats and 6 caps.In how many ways can they wear them?
Number of men = 3
Number of coats = 4
Number of waist coats = 5
Number of caps = 6
4 coats can be given to 3 men in 4P 3 ways.
5 waist coats can be given to 3 men in 5P 3 ways.
6 caps can be given to 3 men in 6P 3 ways.
∴ Total number of ways of wearing 3 coats, 4 waist coats and 6 caps is
= 4P 3 × 5P 3 × 6P 3
= 1, 72, 800
SIMPLE
Total Number of letters = 6
They can be arranged in 6! ways
∴ The number of words = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720
Each question has 4 choices. So each question can be answered in 4 ways.
Number of Questions = 10
So they can be answered in 410 ways
(ii) The first four questions have 3 choices. So they can be answered in 3 4 ways.
The remaining 6 questions have 5 choices. So they can be answered in 5 6 ways.
So all 10 questions can be answered in 3 4 × 5 6 ways.
(iii) Given question n has n + 1 choices
(i) What is the maximum number of different answers can the students give?
Selecting a correct answer from the 4 answers can be done in 4 ways.
Total number of questions = 5 So they can be answered in 45 ways
(ii) How will the answer change if each question may have more than one correct answers? Since each question may have more than one correct answer, each question can have the possibilities 1, 2, 3 or 4 correct answers.
∴ Number of ways of answering each question
= 4C 1 + 4C 2 + 4C 3 + 4C 4
= 4 + 6 + 4 + 1 = 15
Thus, the answer will change as 15 5 (i.e, Total number of ways of answering five questions).
ARTICLE
Vowels A, I, E = 3
Total number of places = 7
1 2 3 4 5 6 7
Number of even places = 3
3 Vowels can occupy 3 places in 3! = 3 × 2 × 1 = 6 ways
Then the remaining 4 letters can be arranged in 4! ways
So total number of arrangement = 3! × 4! = 6 × 24 = 144 ways
Number of women standing in a line = 8
Number of men standing in a line = 6
(i) How many arrangements are possible if any individual can stand In any position?
Total number of persons = 8 + 6 = 14
They can be arranged in 14! ways
(ii) In how many arrangements will all 6 men be standing next to one another?
There are 6 men and 8 women. To make all 6 men together treat them as 1 unit. Now there are 1 + 8 = 9 persons.
They can be arranged in 9! ways. After this arrangement, the 6 men can be arranged in 6! ways.
So total number of arrangements = 9! × 6!
(iii) In how many arrangements will no two men be standing next to one another?
8 women placed in the box can be arranged in 8P 8 = 8! ways
∴ Total number of ways = 9P 6 × 8!
The given word is MISSISSIPPI.
Number of letters in the word = 11
Number of S’s = 4
Number of I’s = 4
Number of P’s = 2
Number of M’s = I
Hence the total number of distinct words
The given term is a 2 b 3 c 4
The factors are a, b, c
Number of a’s = 2
Number of b’s = 3
Number of c’s = 4
a 2 b 3 c 4 = a×a × b×b×b × c×c×c×c
Total number of factors in the product = 9
Number of ways the product can be expressed without exponents
Number of maths book = 4
Number of physics books = 3
Number of chemistry books = 2
Number of biology books = 1
Since we want books of the same subjects together, we have to treat all maths books as 1 unit, all physics books as 1 unit, all chemistry books as 1 unit, and all biology books as 1 unit.
Now the total number of units = 4
They can be arranged in 4! ways. After this arrangement.
4 maths book can be arranged in 4! ways
3 physics book can be arranged in 3! ways
2 chemistry book can be arranged in 2! ways and 1 biology book can be arranged in 1! way
∴ Total Number of arrangements 4! 4! 3! 2! = 6912
SUCCESS
Number of letters = 7
Number of ‘S’ = 3
Since we want all ‘S’ together treat all 3 S’s as 1 unit.
Now the remaining letters = 4
∴ Total number of unit = 5
They can be arranged in 5! ways of them C repeats two times.
So total number of arrangements = \(\frac{5 !}{2 !}\) = 60
(i) How many different sequences of heads and tails are possible?
A coin on tossing has two outcomes.
Tossing a coin once number of outcomes = 2
∴ Tossing a coin 8 times number of outcomes = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2 8
∴ The different sequences of heads and tails are 2 8
(ii) How many different sequences containing six heads and two tails are possible?
When tossing a coin, the possible outcomes are head
and tail, Total number of outcomes of heads and tails
on tossing a coin 8 times is
= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2 8
The number of ways of getting six heads and 2 tails
= 4 × 7 = 28 ways
The given word is INTERMEDIATE
Number of letters = 12
Number of I’S = 2
Number of T’S = 2
Number of E’S = 3
Vowels are A,I, I, E, E, E
Total number of vowels = 6
Consonants are N, T, R, M, D, T
Total number of consonants = 6
(i) Vowels and consonants are alternative
(a) Let the first box be filled with a vowel. There are six alternate places available for 6 vowels.
∴ Number of ways of filling 6 vowels in the alternative six boxes is \(\frac{6 !}{2 ! \times 3 !}\)
Remaining 6 boxes can be filled with the 6 consonants. Number of ways of filling the 6 consonants in the remaining 6 boxes is \(\frac{6 !}{2 !}\)
Total number of ways = \(\frac{6 !}{2 ! \times 3 !} \times \frac{6 !}{2 !}\)
(b) Let the first box be filled with a consonant. There six alternate places available for 6 consonants.
∴ Number of ways of filling 6 consonants in the six alternate boxes is \(\frac{6 !}{2 !}\)
Remaining 6 boxes can be filled with the 6 vowels
∴ Number of ways of filling the 6 vowels in the remaining 6 boxes is \(\frac{6 !}{2 ! \times 3 !}\)
Total number of ways = \(\frac{6 !}{2 !} \times \frac{6 !}{2 ! \times 3 !}\)
∴ Total number of strings formed by using the letters of the word INTERMEDIATE, if the vowels and consants are alternative
(ii) All the vowels are together.
Take all the vowels as 1 unit.
Remaining consonants = 6
Total number of units for arrangement = 7
Number of arrangements = \(\frac{7 !}{2 !}\)
(Since in the consonant’s T’ s repeated twice)
Among the vowels, the number of arrangements
∴ Total number of arrangements = 2520 × 60
= 151200
(iii) Vowels are never together:
Total number of arrangements using all the
Number of arrangements where all the vowels are together = 151200
Number of arrangements where all vowels are never together = Total number of arrangements – number of arrangements where the vowels are together = 19958400 – 151200 = 19807200
(iv) No two vowels are together:
same as (i)
The number of strings using the letters of the word INTERMEDIATE if no two vowels are together is 43,200.
(i) How many distinct 6-digit numbers are there?
Given numbers are 1, 1, 2, 3, 3, 4
Here 1 occurs twice, 3 occurs twice.
∴ Number of distinct 6 – digit numbers
(ii) How many of these 6-digit numbers are even?
The given numbers are 1, 1, 2, 3, 3, 4
In order to get even 6 – digit numbers, the unit place must be filled by the digits 2 or 4. Therefore, the unit place can be filled in 2 ways using the digits 2 or 4. In the remaining 5 digits (excluding the digit placed in the unit place 2 or 4) 1 occurs 2 times, 3 occurs 2 times.
∴ The number of ways of filling other places using
the remaining 5 digit is = \(\frac{5 !}{2 ! \times 2 !}\)
∴ Number of distinct 6 – digit numbers = \(\frac{5 !}{2 ! \times 2 !} \times 2\)
= 5 × 4 × 3
= 60
(iii) How many of these 6 – digit numbers are divisible by 4?
In order to get the 6 – digit number divisible by 4, the last two digits must be divisible by 4
∴ The last two digits should be 12 or 24 or 32
Let the last box be filled with 24. The remaining 4 boxes can be filled with the remaining digits
Let the last box be filled with 12. The remaining 4 boxes can be filled with the remaining digits
Let the last box be filled with 32. The remaining 4 boxes can be filled with the remaining digits
The total number of 6 digit numbers which are divisible by 4 is
= 6 + 12 + 12 = 30
∴ Required number of 6 – digit numbers = 30
(i) GARDEN
The letters of the word arranged in the dictionary order is
A, D, E, G, N, R
Total number of letters = 6
The number of words begins with A = 5!
The number of words begins with D = 5!
The number of words begins with E = 5!
The number of words beginning with G = 5!
(But one of these words is GARDEN)
The number of words beginning with GAD = 3!
The number of words beginning with GAE = 3!
The number of words beginning with GAN = 3!
There are 3 ! words beginning with GAR one of these words is GARDEN. The first word beginning with GAR is the word GARDEN.
∴ The rank of the word GARDEN 3 × 120 + 3 × 6 + 1 = 360 + 18 + 1 = 379
(ii) DANGER
The dictionary order of the letters of the given word is A, D, E, G, N, R
In the dictionary order of words which begin with A, comes first. If we fill the first place with A, the remaining 5 letters can be arranged in 5! ways. Proceeding like this
Number of words beginning with D = 5! = 120
Number of words beginning with DAE = 3! = 6
Number of words beginning with DAG = 3! = 6
Number of words beginning with DANE = 2! = 2
Number of words beginning with DANGE = 1! = 1
(which is the word DANGER)
∴ The rank of the word DANGER = 120 + 6 + 6 + 2 + 1 = 135
The given word is THING
Arranging the letters of the word in the dictionary order, we have G, H, I, N, T
The number of strings that can be made using all the letters T, H, I, N, G of the word THING is = 5! = 120
The number of words beginning with G = 4!
The number of words beginning with H = 4!
The number of words beginning with I = 4!
Number of words so far formed = 4! + 4! + 4!
= 24 + 24 + 24 = 72 words
As the required word is in the 85 th position, the required word must begin with N Number of words beginning with NG = 3!
A number of words beginning with NH = 3!
Total number of words so far formed
= 72 + 3! + 3!
= 72 + 6 + 6
= 84 words
The next string is the required string. It should begin with NI and its first word beginning with NI which is NIGHT
∴ 85 th strings are NIGHT
The given word is FUNNY
To find the rank of the word FUNNY, write down the letters of the word FUNNY. Other than F in alphabetical order N, N, U, Y
Number of words beginning with F = \(\frac{4 !}{2 !}\)
Among these words, the number of words beginning with FN = 3! = 1 × 2 × 3 = 6words
(Treating FN as one unit, the remaining 3 letters can be arranged in 3 ! ways)
The number of words beginning with FU is = \(\frac{3 !}{2 !}\)
Among these three words, FUNNY is the first word, hence among the twelve words beginning with F, FUNNY appears as the 7 th word.
The given numbers are 1, 2, 3, 4, 5
The total number of arrangements. Using the digits 1, 2, 3, 4 and 5 taking 4 at a time is 5 P 4
= 5 × 4 × 3 × 2 = 120
∴ 120 four-digit numbers can be formed using the given 5 digits without repetition. To find the sum of these numbers, we will find the sum of digits at unit’s, ten’s, hundred’s and thousand’s place in all these 120 numbers.
Consider the digit in unit’s place. In all these numbers, each of these digits 1, 2, 3, 4, 5 occurs 120
in \(\frac{120}{5}\) = 24 times in the units place.
∴ The sum of the digits at unit’s place
= 24(1 + 2 + 3 + 4 + 5)
= 24 × 15 = 360
Similarly sum of the digit’s at ten’s place = 360
Sum of the digit’s at hundred’s place = 360
Sum of the digit’s at thousand’s place = 360
∴ Sum of all four digit numbers formed using the digit’s 1, 2, 3, 4, 5
= 360 × 10° + 360 × 10 1 + 360 × 10 2 + 360 × 10 3
= 360 (10° + 10 1 + 10 2 + 10 3 )
= 360 (1 + 10 + 100 + 1000)
= 360 × 1111
= 3,99,960
The given digits are 0, 2, 5, 7, 8
The first box can be filled in 4 ways using the digits 2, 5, 7, 8 ( excluding 0). The second box can be filled in 4 ways using the digits 0, 2, 5, 7, 8 excluding the digit placed in the first box. The third box can be filled in 3 ways using the digits 0, 2, 5, 7, 8 excluding the digits placed in the first two boxes. The fourth box can be filled in 2 ways using the digits 0, 2, 5, 7, 8 excluding the digits placed in the first three boxes.
∴ Total number of 4-digit numbers = 4 × 4 × 3 × 2 = 96
To find the sum of all these four-digit numbers.
Fix the number 0 in the list box (4). With the remaining numbers 2, 5, 7, 8, box – 3 can be filled in 4 ways, box – 2 can be filled in 3 ways, and box – 1 can be filled in 2 ways.
∴ Total number of 4 digit numbers ending with 0 is = 4 × 3 × 2 = 24 numbers
Fix the number 2 in the last box -4. With the remaining digits 0, 5, 7, 8. Box -l can be filled in 3 ways excluding the digit 0. Box – 2 can be filled in 3 ways using the digits 0, 5, 7, 8 excluding the digit placed in a box – 1. Box – 3 can be filled in 2 ways using the digits 0, 5, 7, 8 excluding the digits placed in box-1 and box-2.
∴ Total number of 4 – digit numbers ending with the digit 2 = 3 × 3 × 2 = 18 numbers
Similarly, Total numbers of 4 – digit numbers ending with the digit 5 = 18 numbers
Total number of 4 – digit numbers ending with the digit 7 = 18 numbers
Total number of 4 – digit numbers ending with the digit 8 = 18 numbers
∴ Total for unit place = (24 × 0) + (18 × 2) + (18× 5) + ( 18 × 7) +( 18 × 8)
= 18 × (2+ 5 + 7 + 8)
= 18 × 22 = 396
∴ Sum of the digits at the unit place = 396
Similarly Sum of the digits at ten’s place = 396
Sum of the digit’s at hundred’s place = 396
Sum of the digit’s at thousand’s place = 396
∴ Sum of all four digit numbers formed using the digits 0, 2, 5, 7, 8
= 396 × 10° + 396 × 10 1 + 396 × 10 2 + 396 × 10 3
= 396 × ( 10° + 101 + 102 + 103)
= 396 × ( 1 + 10 + 100 + 1000)
= 396 × 1111 = 571956
Given nC 12 = nC 9
We have nC x = nC y ⇒ x = y or x + y = n
nC 12 = nC 9
⇒ 12 + 9 = n ⇒ n = 21
⇒ 21C n = 21C 21 = 1
Given 15C 2r – 1 = 15C 2r + 4
We have nC x = nC y ⇒ x = y or x + y = n
15C 2r – 1 = 15C 2r + 4
⇒ (2r – 1) + (2r + 4) = 15
4r + 3 = 15 ⇒ 4r = 12
r = \(\frac{12}{4}\) = 3
Given nP r = 720, If nC r = 120
r ! = 6
r ! = 3 × 2 × 1 = 3!
r = 3
n (n – 1) (n – 2) = 720
n(n – 1) (n – 2) = 1o × 9 × 8
n = 10
∴ r = 3, n = 10
(n + 1) n(n – 1)(n – 2) = 57 × 7 × 6 × 5 × 4 × 3
= 3 × 19 × 7 × 6 × 5 × 4 × 3
= (7 × 3) × (5 × 4)19 × (6 × 3)
= 21 × 20 × 19 × 18
= (20 + 1) × (20) × (20 – 1) × (20 – 2)
∴ n = 20
Number of kabaddi players = 14
7 players must be selected from 14 players
Number of ways of selecting 7 players from 14 players is
(ii) There are 15 persons in a party and if each 2 of them shakes hands with each other, how many handshakes happen in the party?
Total number of person in the party = 15
Given if each 2 of the 15 persons shakes bands with each other.
∴ The total number of hand shakes is same as the number of ways of selecting 2 persons among 15 persons. This can be done in 15C 2 ways.
Number of hand shakes = 15C 2
(iii) How many chords can be drawn through 20 points on a circle ?
Number given points on the circle = 20
A chord is obtained by joining any two points on the circle.‘Number of chords drawn through 20 points is same as the number of ways of selecting 2 points out of 20 points. This can be done 20C 2 ways.
∴ The total number of chords = 20C 2
(iv) In a parking lot one hundred, one year old cars are parked. Out of them five are to be chosen at random for to cheek its pollution devices. How many different set of five cars can be chosen?
Number of cars in the parking lot = 100
Number of cars to be selected to check pollution device = 5
Number of ways of selecting 5 cars out of 100 cars
(v) How many ways can a team of 3 boys, 2 girls and 1 transgender be selected from 5 boys, 4 girls and 2 transgenders ?
Number of boys = 5
Number of girls = 4
Number of transgender = 2
Number of ways of selecting 3 boys from 5 boys = 5c 3
Number of ways of selecting 2 girls from 4 girls = 4C 2
Number of ways of selecting one transgender from 2 transgenders = 2C 1
Total number of ways of selection
(i) Subsets with 4 elements
Number of subsets with no element = 4C o
Number of subsets with one element = 4C 1
Number of subsets with two elements = 4C 2
Number of subsets with three elements = 4C 3
Number of subsets with four elements = 4C 4
∴ Total number of subsets
(ii) Subsets with 5 elements:
Number of subsets with no element = 5C 0
Number of subsets with one element = 5C 1
Number of subsets with 2 elements = 5 C 2
Number of subsets with 3 elements = 5C 3
Number of subjects with 4 elements = 5C 4
Number of subsets with 5 elements = 5C 5
Total number of subjects
(iii) Subsets with n elements
Number of subsets with no element = nC 0
Number of subsets with 1, 2, 3, 4, …………. n elements are nC 1, nC 2, nC 3, nC 4 …………… nC n respectively.
∴ Total number of subjects
= nC 0 + nC 1 + nC 2 + nC 3 + ………… + nC n
= Sum of the coefficients in the binomial expansion (x + a) n
= 2 n
Number of members in the trust = 25
(i) How many ways 3 officers can be selected?
The number of ways of selecting 3 officers from 25 members is
(ii) In how many ways can a president, vice president, and secretary be selected?
From the 25 members, a president can be selected in 25 ways
After the president is selected, 24 persons are left out.
So a Vice President can be selected in (from 24 persons) 24 ways.
After the selection of Vice President, 23 persons are left out
So a secretary can be selected (from the remaining 23 persons) in 23 ways
So a president, Vice president, and a secretary can be selected in 25 P 3 ways
25 P 3 = 25 × 24 × 23 = 13800 ways
Number of persons = 10
Number of committee members to be selected = 6
A committee must have a chairperson and a secretary, the remaining 4 members.
The number of ways of selecting a chairperson and a secretary from 10 persons is
After the selection of chairperson and secretary remaining number of persons = 8
Number of ways of selecting remaining 4 committee members from the remaining 8 persons
= 2 × 7 × 5 = 70
∴ A number of ways of selecting the six committee members from 10 persons.
= 45 × 70 = 3150 ways
Total number of books = 12
Number of books to be selected = 5
Given Two books are always selected.
Remaining number of books to be selected = 3
The number of ways of selecting the remaining 3 books from the remaining 10 books = 10C 3
(ii) Two particular books are never selected. Since two books are never selected, the total number of books is 10.
∴ The number of ways of selecting 5 books from 10 books
= 2 × 9 × 2 × 7 = 252 ways
The number of teachers = 5
Number of students = 20
The number of ways of selecting 2 teachers from 5 teachers is = 5C 2 ways.
The number of ways of selecting 3 students from 20 students is = 20C 3
= 20 × 19 × 3 = 1140 ways
∴ The total number of selection of the committees with 2 teachers and 5 students is = 10 × 1140 = 11400
(i) A particular teacher is included.
Given a particular teacher is selected. Therefore, the remaining 1 teacher is selected from the remaining 4 teachers. Therefore, the number of ways of selecting 1 teacher from the remaining 4 teacher = 4C 1 ways = 4 ways
The number of ways of selecting 3 students from 20 students = 20C 3
Hence the required number of committees
= 1 × 4 × 1140 = 4560
(ii) A particular student is excluded The number of ways of selecting 2 teachers from 5 teachers is = 5C 2
A particular student is excluded
∴ The number of remaining students = 19
Number of ways of selecting 3 students from 19 students
= 19 × 3 × 17 = 969
∴ The required number of committees = 10 × 969 = 9690
Total number of questions = 9
Number of questions to be answered = 5
Since 2 questions are compulsory, a student must select 3 questions from dying remaining 7 questions* The number of ways of selecting 3 questions from 7 questions is = 7C 3
∴ The number of ways of answering 5 questions = 7C 3
Total number of cards in a pack 52
Number of aces = 4
Number of cards to be selected = 5
The number of ways of selecting 3 aces from 4 aces is 4C 3
The number of ways of selecting the remaining 2 cards
from the remaining 48 cards (52 – 4 aces cards) = 48C 2
∴ Required number of ways of selection
= 4C 3 × 48C 2
= 4C 1 × 48C 2
= 4 × 24 × 47 = 4512
Number of Indians = 7
Number of Americans = 5
Number of members in the committee = 5
Selection of 5 members committee with majority Indians
Case (i) 3 Indians and 2 Americans
The number of ways of selecting 3 Indians from 7 Indians is = 7C 3
The number of ways of selecting 2 Americans from 5 Americans is = 5C 2
Total number of ways in this case is = 7C 3 × 5C 2
Case (ii) 4 Indians and 1 American
The number of ways of selecting 4 Indians from 7 Indians is = 7C 4
The number of ways of selecting I American from 5 Americans is = 5C 1
The total number of ways, in this case, is = 7C 4 × 5C 1
Case (iii) 5 Indians no American
Number of ways of selecting 5 Indians from
7lndiansis = 7C 5
Total number of ways, in this case, = 7C 5 × 5C 0
∴ Total number of ways of forming the committee
= 7 × 5 × 5 × 24 – 7 × 5 × 5 + 7 × 3
= 350 + 175 + 21 = 546
Number of men = 8
Number of women = 4
Number of peoples in the committee = 7
(i) Exactly 3 women
In a 7 member committee, women must be 3. Therefore, the remaining 4 must be men.
The number of ways of selecting 3 women from 4 women = 4C 3
The number of ways of selecting 4 men from 8 men = 8C 4
∴ The total number of ways of selection is = 4C 3 × 8C 4
= 8 × 7 × 5 = 280
(ii) At least 3 women
The 7 members committee must contain atleast 3 women
∴ We have the following possibilities
(i) 4 women + 3 men
(ii) 3 women + 4 men
case (i) 4 women + 3 men
The number ways of selecting 4 women.from
4 women is = 4C 4 = 1 way
The number of ways of selecting 3 men from 8 men = 8C 3
∴ The total number of ways = 1 × 56 = 56
case (ii) 3 women + 4 men
The number of ways of selecting 3 women from 4 women is = 4C 3
The number of ways of selecting 4 men from 8 men is = 8C 4
∴ The total number of ways = 4C 3 × 8C 4
= 8 × 7 × 5 = 280
∴ The required number of ways of forming the committee = 56 + 280 = 336
(iii) atmost 3 women
The 7 members must contain at most 3 women,
∴ we have the following possibilities
(i) No women + 7 men
(ii) 1 women + 6 men
(iii) 2 women + men
(iv) 3 women + 4 men
Case (i) 0 women + 7 men
The number of ways of selecting O women from 4 women is = 4C 0
The number of ways of selecting 7 men from 8 men is = 8C 7
Total number of ways = 4C 0 × 8C 7
Case (ii) 1 women + 6 men
The number of ways of selecting 1 woman from 4 women is = 4C 1
The number of ways of selecting 6 men from 8 men is = 8C 6
Total number of ways = 4C 1 × 8C 6
Case (iii) 2 women + 5 men
The number of ways of selecting 2 women from 4 women is = 4C 3
The number of ways of selecting 4 men from 8 men is = 8C 4
∴ Total number of ways = 4C 3 × 8C 4
∴ The required number of ways of forming the committee
Number of relatives of a man = 7
4 ladies and 3 gentlemen
Number of relatives of the man’s wife = 7
3 ladies and 4 gentlemen
The dinner party consists of 3 ladies and 3 gentlemen.
For the dinner party, 3 persons from the man’s relatives and 3 persons from the man’s wife’s relatives are invited.
Then we have the following possibilities for the different possible arrangements.
Required number of ways
Number of white balls = 2
Number of black balls = 3
Number of red balls = 4
Number of balls drawn = 3 balls with atleast 1 blackball
We have the following possibilities
Required number of ways of drawing 3 balls with atleast one black ball.
There are 11 fetters in the word EXAMINATION,
They are AA, II, NN, E, X, M, T, O The four-letter strings may have
(i) 2 alike letters of one kind and 2 alike tetters of the second kind.
(ii) 2 alike fetters and 2 different letters.
(iii) All different fetters.
(i) 2 alike letter of one kind and 2 alike letters of the second kind:
There are three sets of 2 alike letters AA, II, NN. Out of these sets, two sets can be selected in 3C 2 ways. So there are 3C 2 groups each of which còntains 4 letter
strings out of which 2 are alike of one kind type and 2 are alike of the second type.
4 letters in each group can be arranged in \(\frac{4 !}{2 ! \times 2 !}\) ways.
Hence the total number of strings consisting of two alike letters of one kind and 2 alike letters of the second kind
(ii) 2 alike letter and 2 different letters:
Out of sets of two alike letters, one set can be chosen in 3C 1 ways. From the remaining 7 distinct letters, 2 letters can be chosen in 7C 2 ways. Thus 2 alike letters and 2 different letters can be selected in (3C 1 × 7C 2 ) ways. There are ( 3C 1 × 7C 2 ) groups of 4 letters each. Now letters of each group can be arranged among themselves in \(\frac{4 !}{2 !}\) ways.
Hence the total number of strings consisting of 2 alike and 2 distinct letters,
(iii) All different letters
There are 8 different letters
E, X, A, M, I, N, T, O
Out of which 4 can be selected in 8C 4 ways. So there are 8C 4 groups of 4 letters each The letter in each of 8C 4 groups’ can be arranged in 4! ways.
∴ The total number of 4. letter strings in which all letters are distinct = 8C4 × 4!
= 56 × 30 = 1680 strings
Hence the total number of 4 letter strings
= 18 +756 + 1680
= 2454 strings
Number of points 15
To form a triangle we need 3 non – collinear points.
The number of ways of se1ecting 3 non-collinear points from 15 points is = 15C 3
To form a triangle we need 3 non-collinear points. Take the 7 points lying on one line be group A and the remaining 8 points lying on another parallel line be group B.
We have the following possibilities
∴ Required number of ways of forming the triangle
Number of points in a plane = 11
No three of these points lie in the same straight line except 4 points.
(i) The number of straight lines that can be obtained from the pairs of these points
Through any two points, we can draw a straight line.
∴The number straight lines through any two points of the given 11 points = 11C 2
Given that 4 points are collinear. The number of straight lines through any two points of these 4 points is
Since these 4 points are collinear, these 4 points contribute only one line instead of the 6 lines.
The total number of straight lines that can be drawn through 11 points on a plane with 4 of the points being collinear is = 55 – 6 + 1 = 50
(ii) The number of triangles that can be formed for which the points are their vertices.
To form a triangle we need 3 non – collinear points. We have the following possibilities.
(a) If we take one point from 4 collinear points and 2 from the remaining 7 points and join them.
The number of ways of selecting one point from the 4 collinear points is = 4C 1 ways
The number of ways of selecting 2 points from the remaining 7 points is = 7C 2
The total number of triangles obtained in this case is = 4C 1 × 7C 2
∴ The total number of triangles obtained in this case is
(b) If we select two points from the 4 collinear points and one point from the remaining 7 points then the number of triangles formed is
(c) If we select all the three points from the 7 points then the number of triangles formed is
∴ The total number of triangles formed are
= 84 + 42 + 35 = 161
Let the number sides of the polygon be n.
The number of diagonals of a polygon with n sides = \(\frac{n(n-3)}{2}\)
Given \(\frac{n(n-3)}{2}\) = 90
n 2 – 3n = 180
n 2 – 3n = 180
n 2 – 3n – 180 = 0
n 2 – 15n + 12n – 180 = 0
n (n – 15 ) + 12 (n – 15) = 0
(n + 12) (n – 15) = 0
n + 12 = 0 or n – 15 = 0
n = – 12 or n = 15
But n = – 12 is not possible
∴ n = 15
∴ The number of sides of the polygon having 90 diagonals is 15.
Let P(n) = 1 2 + 2 2 + 3 2 + ……….. + n 3
p(n) = \(\left(\frac{\mathbf{n}(\mathbf{n}+\mathbf{1})}{2}\right)^{2}\)
Step 1:
Let us verify the statement for n = 1
p(1) = 1 3 = \(\left(\frac{1(1+1)}{2}\right)^{2}\)
P(1) = 1
∴ The given statement is true for n = 1
Step 2:
Let us assume that the given statement is true for n = k
P(k) = 1 3 + 2 3 + 3 3 + …………… + k 3
Step 3:
Let us prove the statement is true for n = k + 1
P(k + 1 ) = 1 3 + 2 3 + 3 3 + …………… + k 3 + (k + 1) 3
This implies P (k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.
1 2 + 2 2 + 3 2 + ……….. + n 3 = \(\left(\frac{\mathbf{n}(\mathbf{n}+\mathbf{1})}{2}\right)^{2}\)
is true for all natural numbers n.
Step 1:
Let us verify the statement for n = 1
∴ The given statement is true for n = 1
Step 2:
Let us assume that the given statement is true for n = k
Step 3:
Let us prove the statement is true for n = k + 1
This implies P (k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.
is true for all natural numbers n.
Let P(n): 2 + 4 + 6 +…+2n = n 2 + n, ∀ n ∈ N
Step 1:
P( 1): 2 = 1 2 + 1 = 2 which is true for P( 1)
Step 2:
P(k): 2 + 4 + 6+ …+ 2k = k 2 + k. Let it be true.
Step 3:
P(k + 1): 2 + 4 + 6 + … + 2k + (2k + 2)
= k 2 + k + (2k + 2) = k 2 + 3k + 2
= k 2 + 2k + k + 1 + 1
= (k+ 1) 2 + (k + 1)
Which is true for P(k + 1)
So, P(k + 1) is true whenever P(k) is true.
Let P(n) = 1. 2 + 2. 3 + 3. 4 + ………… + n. (n + 1) = \(\frac{\mathbf{n}(\mathbf{n}+1)(\mathbf{n}+2)}{3}\)
Step 1:
Let us verify the statement for n = 1
P(1) = 1. 2 = \(\frac{1(1+1)(1+2)}{3}\)
= 1. 2 = \(\frac{1 \cdot 2 \cdot 3}{3}\)
∴ The result is true for n = 1
Step 2:
Let us assume that the result is true for n = k
P (k) = 1-2 + 2-3 + 3- 4 + + k (k + 1 ) =
Step 3:
Let us prove the result for n = k + 1
P(k + 1) = 1. 2 + 2. 3 + 3. 4 + …………… + k (k + 1) (k + 1) (k + 1 + 1)
P (k + 1) = 1. 2 + 2. 3 + 3. 4 + ………….. + k (k + 1) + (k + 1 ) (k + 2)
P (k + 1 ) = P(k) + (k + 1) (k + 2)
This implies P (k + 1) is true.
∴ Thus, we have proved the result for n = k + 1, Hence by the principle of mathematical induction, the result is true for all natural numbers n.
1. 2 + 2. 3 + 3. 4 + ………. + n (n + 1) = \(\frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}{3}\)
is true for all natural numbers n ≥ 1
Let
The first stage is n = 2
Step 1:
Let us verify the result for n = 2
∴ The given result is true for n = 2.
Step 2:
Let us assume that the result is true for n = k.
Step 3:
Let us prove the result for n = k + 1
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n ≥ 2.
is true for all natural numbers n ≥ 2.
Step 1:
Since n ≥ 2 the first stage is n = 2
Let us verify the result for n = 2
∴ P (2 ) is true.
The result is true for n = 2.
Step 2:
Let us assume that the result is true for n = k.
Step 3:
Let us prove the result for n = k + 1
This implies p(k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n ≥ 2.
is true for all natural numbers n ≥ 2
Step 1:
First, let us verify the result for n = 1
∴ The given result is true for n = 1
Step 2:
Let us assume the result for n = k
Step 3:
Let us prove the result for n = k + 1
Factorizing k 3 + 6k 2 + 9k + 4
f (k) = k 3 + 6k 2 + 9k + 4
f(- 1) = (- 1) 3 + 6(- 1) 2 + 9(- 1) + 4
f(- 1) = – 1 + 6 – 9 + 4 = 0
∴ (k + 1) is a factor of f(k)
k 3 + 6k 2 + 9k + 4 = (k + 1) (k 2 + 5k + 4).
= (k + 1) (k 2 + 4k + k + 4)
= (k + 1) [k(k + 4) + 1(k + 4)]
= (k + 1) (k + 4) (k + 1)
This implies P(k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.
is true for all natural numbers n
Step 1:
First, let us verify the result for n = 1
∴ The result is true for n = 1
Step 2:
Let us assume that the result is true for n = k
Step 3:
Let us prove the result for n = k + 1
This implies P(k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.
P(n) is the statement
1! + (2 × 2!) + (3 × 3!) + ….. + (n × n!) = (n + 1)! – 1
To prove for n = 1
LHS = 1! = 1
RHS = (1 + 1)! – 1 = 2! – 1 = 2 – 1 = 1
LHS = RHS ⇒ P(1) is true
Assume that the given statement is true for n = k
(i.e.) 1! + (2 × 2!) + (3 × 3!) + … + (k × k!) = (k + 1)! – 1 is true
To prove P(k + 1) is true
p(k + 1) = p(k) + t (k + 1)
P(k + 1) = (k + 1)! – 1 + (k + 1) × (k + 1)!
= (k + 1)! + (k + 1) (k + 1)! – 1
= (k + 1)! [1 + k + 1] – 1
= (k + 1)! (k + 2) – 1
= (k + 2)! – 1
= (k + 1 + 1)! – 1
∴ P(k + 1) is true
⇒ P(k) is true, So by the principle of mathematical induction P(n) is true.
Let P(n) = x 2n – y 2n is divisible by (x + y)
For n = 1
P(1) = x 2 × 1 – y 2 × 1 is divisible by (x + y)
⇒ (x + y) (x – y) is divisible by (x + y)
∴ P(1) is true
Let P(n) be true for n = k
∴ P(k) = x 2k – y 2k is divisible by (x + y)
⇒ x 2k – y 2k = λ(x + y) …… (i)
For n = k + 1
⇒ P(k + 1) = x 2(k + 1) – y 2(k + 1) is divisible by (x + y)
Now x 2(k + 2) – y 2(k + 2)
= x 2k + 2 – x 2k y 2 + x 2k y 2 – y 2k + 2
= x 2k.x 2 – x 2k y 2 + x 2k y 2 – y 2k y 2
= x 2k (x 2 – y 2 ) + y 2 λ (x + y) [Using (i)]
⇒ x 2k + 2 – y 2k + 2 is divisible by (x + y)
∴ P(k + 1) is true.
Thus P(k) is true ⇒ P(k + 1) is true.
Hence by the principle of mathematical induction, P(n) is true for all n ∈ N
Let P (n) = 1 2 + 2 2 + 3 2 + ………….. + n 2 > \(\frac{n^{3}}{3}\)
Step 1:
Let us first verify the result for n = 1
∴ The result is true for n = 1
Step 2:
Let us assume that the result is true for n = k.
P(k) = 1 2 + 2 2 + 3 2 + …………… + k 2 > \(\frac{\mathrm{k}^{3}}{3}\)
Step 3:
Let us prove the result for n = k + 1
P(k + 1) = 1 2 + 2 2 + 3 2 + …………… + k 2 + (k + 1) 2
P(k + 1) = P(k) + ( k + 1) 2
This implies P (k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.
1 2 + 2 2 + 3 2 + ………….. + n 2 > \(\frac{n^{3}}{3}\)
Let P(n): n 3 – 7n + 3
Step 1:
P(1) = (1) 3 – 7(1) + 3
= 1 – 7 + 3 = -3 which is divisible by 3
So, it is true for P(1).
Step 2:
P(k): k 3 – 7k + 3 = 3λ. Let it be true
⇒ k 3 = 3λ + 7k – 3
Step 3:
P(k + 1) = (k + 1) 3 – 7(k + 1) + 3
= k 3 + 1 + 3k 2 + 3k – 7k – 7 + 3
= k 3 + 3k 2 – 4k – 3
= (3λ + 7k – 3) + 3k 2 – 4k – 3 (from Step 2)
= 3k 2 + 3k + 3λ – 6
= 3(k 2 + k + λ – 2) which is divisible by 3.
So it is true for P(k + 1).
Hence, P(k + 1) is true whenever it is true for P(k).
To prove 5 n + 1 + 4 × 6 n when divided by 20 leaves a remainder 9
That is to prove 5 n + 1 + 4 × 6 n – 9 is divisible by 20
Let P (n) be the statement 5 n + 1 + 4 × 6 n
which is divisible by 20
P (n) = 5 n + 1 + 4 × 6 n – 9 = 20λ
Step 1:
Let us verify the statement for n = 1
P (1) = 5 i + 1 + 4 × 6 1 – 9
= 5 2 + 4 × 6 – 9
= 25 + 24 – 9
= 49 – 9 = 40
which is divisible by 20
∴ The statement is true for n = 1
Step 2:
Let us assume that the statement is true for n = k
P(k) = 5 k + 1 + 4 × 6 k – 9 is divisible by 20
P(k) = 5 k + 1 + 4 × 6 k – 9 = 20λ …………… (1)
Step 3:
Let us prove the result for n = k + 1
Hence, P(k + 1) is divisible by 20.
This implies P (k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.
Thus, 5 n + 1 + 4 × 6 n when divided by 20 leaves a remainder 9, for all natural numbers n.
Let P( n) = 10 n + 3 × 4 n + 2 + 5 is divisible by 9
Step 1:
Let us verify the result for n = 1
P ( 1 ) = 10 1 + 3 × 4 1 + 2 + 5
= 15 + 3 × 4 3
= 15 + 3 × 64
= 15 + 192 = 207
which is divisible by 9
Thus we have verified the result for n = 1
Step 2:
Let us assume the result is true for n = k
P(k) = 10 k + 3 × 4 k + 2 + 5 is divisible by 9
∴ 10 k + 3 × 4 k + 2 + 5 = 9m for some m
10 k = 9m – 5 – 3 × 4 k + 2 for some m
Step 3:
Let us prove the result for n = k + 1
which is divisible by 9
This implies P (k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.
10 n + 3 × 4 n + 2 + 5 is divisible by 9,
for all natural numbers n.
This implies P (k + 1) is true. Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.
(2) 108
Explaination:
Number of digits given = 2, 4, 5, 7
Number of 4 digit numbers formed = 4! =24
So each digit occur \(\frac{24}{4}\) = 6 times
Sum of the digits = 2 + 4 + 5 + 7 = 18
So sum of the digits in each place = 18 × 6 = 108
(2) 124
Explanation:
Number of multiple-choice questions = 3
Number of choices in each question = 5
Number of ways of answering = 5 × 5 × 5 = 5 3 = 125
Number of ways of getting all answer correct = 1
∴ Number of ways of getting incorrect answer = 125 – 1 = 124
(1) 30 4 × 29 2
Explanation:
I and II in maths can be given in 30 × 29 ways.
I and II in physics can be given in 30 × 29 ways.
I and chemistry can be given in 30 ways.
I in English can be given in 30 ways.
So total number of ways = 30 × 29 × 30 × 29 × 30 × 30 = 30 4 × 29 2
(2) 5 5
Explanation:
The odd digit numbers = 1, 3, 5, 7, 9
Since repetition is allowed, each digit can be filled by 1, 3, 5, 7, 9
Unit’s place can be filled in 5 ways
Ten’s place can be filled in 5 ways etc.
∴ The number of 5 digit numbers with an odd digit in each place = 5 × 5 × 5 × 5 × 5 = 5 5 ways
(4) 64
Explanation:
Each letter can be ported in 3 ways
∴ 4 letter is 3 4 ways
(2) 6 and 7
Explaination:
( n + 5 ) ( n + 4) = 2 × 11 × (n – 1)
n 2 + 5n + 4n + 20 = 22n – 22
n 2 + 9n + 20 – 22n + 22 = 0
n 2 – 13n + 42 = 0
n (n – 6) – 7 (n – 6) = 0
(n – 7) (n – 6) = 0
n – 7 = 0 or n – 6 = 0
n = 7 and n = 6
(1) r !
Explanation:
1(2) (3) ….. (r) = r! which is ÷ by r!
(4) 69760
Explaination:
Case (i) When zero is allowed in the first place.
The number of five-digit telephone numbers which can
be formed using the digits 0, 1, 2, …………,9 is 10 5.
The number of five-digit telephone numbers which have none of their digits repeated is 10P 5
= 10 × 9 × 8 × 7 × 6 = 30240
∴ The required number of telephone number
= 10 5 – 30240
= 100000 – 30240
= 69,760
Case (ii) When zero is not allowed in the first place,
(a) Repetition allowed:
The number of ways of filling the first box, second box, …………….., fifth box
= 9 × 10 × 10 × 10 × 10
= 90000
(b) Repetition riot allowed:
Number of ways of filling 1 box = 9
Number of ways of filling 2 nd box = 9
Number of ways of filling 3 rd box = 8
Number of ways of filling 4 th box = 7
Number of ways of filling 5 th box = 6
∴ Number of numbers with no digit is repeated = 9 × 9 × 8 × 7 × 6 = 37216
∴ Number of numbers having atleast one of the digits repeated = 90000 – 37216 = 52,784
(2) 3
Explaination:
(a 2 – a)C 2 = (a 2 – a) C 4
(a 2 – a)C 2 = (a 2 – a) C (a 2 – a) – 4 [nC r = nC n-r ]
(a 2 – a)C 2 = (a 2 – a) C a 2 – a – 4
∴ 2 = a 2 – a – 4
a 2 – a – 4 – 2 = 0 ⇒ a 2 – a – 6 = 0
a 2 – 3a + 2a – 6 = 0
a (a – 3) + 2 (a – 3) = 0
⇒ (a + 2)(a – 3) = 0 ⇒ a = – 2 or 3
(2) 40
Explaination:
Total number of points = 10
The number of straight lines joining any two points is same as the number ways of selecting 2 points from 10 points.
∴ The number of straight lines = 10C 2
Given 4 points are collinear.
∴ The number of straight lines that these 4 points contribute = 4C 2
Since 4 points are collinear they lie on a line.
∴ Required number of lines = 10C 2 – 4C 2 + 1
= 5 × 9 – 2 × 3 + 1
= 45 – 6 + 1 = 40
(3) 12C 8 – 10C 6
Explaination:
Number of people = 12
Number of people invited for the party = 8
Number of people not willing to attend the party = 2
Number of ways of selecting 8 people from 12 people is 12C 8
Two people do not attend party out of 10, remaining people 8 can attend in 10C 6 ways.
∴ Number of ways in which two of them do not attend together = 12C 8 – 10C 6
(4) 18
Explaination:
Number of first set of parallel lines = 4
Number of second set of parallel lines = 3
Number of parallelograms = 4C 2 × 3C 2
= \(\frac{4 \times 3}{1 \times 2} \times \frac{3 \times 2}{1 \times 2}\)
= 6 × 3
= 18
(2) 12
Explaination:
Number of shake hands = 66
Let the number of persons = n
First person shakes hands with the remaining n – 1 persons.
Second person shakes hands with the remaining n – 2 persons.
∴ Total number of shake hands
= (n – 1) + (n – 2) + …………. + 2 + 1
= \(\frac{(\mathrm{n}-1)(\mathrm{n}-1+1)}{2}\)
Given 66 = \(\frac{(\mathrm{n}-1) \mathrm{n}}{2}\)
n 2 – n = 132
n 2 – n – 132 = 0
(n – 12) (n + 11) = 0
n = 12 or n = – 11
n = – 11 is not possible.
∴ n = 12
(3) 11
Explaination:
Number of diagonals of a polygon with n sides
= nC 2 – n
= \(\frac{n(n-1)}{2}\)
Given \(\frac{n(n-1)}{2}\) – n = 44
n 2 – n + 2n = 88
n 2 – 3n – 88 = 0
(n – 11 ) (n + 8) = 0
n = 11 and n = – 8
n = – 8 is not possible.
∴ n = 11
(1) 45
Explaination:
Number of non-parallel lines = 10
Two lines will interest.
∴ Number of points of intersection
= 10C 2 = \(\frac{10 \times 9}{1 \times 2}\) = 45
(4) 116
Explaination:
Number of points = 10
Number of triangles formed by using 10 points is same as number of ways of choosing 3 points from 10 points = 10C 3
Also given 4 points are collinear.
∴ These 4 points do not contribute triangles.
Thus, total number of triangles = 10C 3 – 4C 3
= \(\frac{10 \times 9 \times 8}{1 \times 2 \times 3}\) – 4
= 5 × 3 × 8 – 4
= 120 – 4
= 116
(2) 6
Explaination:
Given 2nC 3: nC 3 = 11:
4(2n – 1) = 11 (n – 2)
8n – 4 = 11 n – 22
-4 + 22 = 11 n – 8n
18 = 3n
⇒ n = 6
(3) nC r
Explaination:
[nC r + nC r – 1 = (n + 1 ) C r
(n – 1)C r + (n – 1)C r – 1 = (n – 1 + 1) C r
= nC r
(4) 52C 5 – 48C 5
Explaination:
Number of cards = 52,
Number of kings = 4
Number of ways of choosing 5 cards out of 52 cards = 52 C 5
Number of cards without kings = 48
Number of ways of choosing 5 cards with no kings from 48 cards = 48 C 5
∴ Number of ways of choosing 5 cards out of 52 cards
= 52C 5 – 48C 5
(3) 1296
Explaination:
Number of rectangles in a chessboard is
= 36 × 36
= 1296
(2) 2 10
Explaination:
Each box can be filled in two ways using digits 2 and 3.
∴ A number of 10 digit number that can be written by using the digits 2 and 3 is 2 × 2 × 2 × …………. 10 times. = 210
(1) P n + 1
Explaination:
Given P, = rP r = r!
1 + 1 × P 1 ! + 2 × P 2 + 3 × P 3 + …………… + n × P n !
= 1 + (1 × 1! + 2 × 2! + 3 × 3! + …………. + n × n!)
= (n + 1 )!
= P n + 1
[(1 × 1 !) + (2 × 2!) + (3× 3 !) + ….. + (n × n !) = (n + 1)! – 1]
(2) \(\left(\frac{1}{2}\right)^{n}\) 2nC n × nP n
Explaination:
1. 3. 5 …………. (2n – 1)
(1) 14
Explaination:
Given nC n, nC 5, nC 6 are in A. P.
∴ 2 × nC 5 = nC 4 + nC 6
12(n – 4) = 30 + (n – 4) (n – 5)
12n – 48 = 30 + n 2 – 5n – 4n + 20
12n – 48 = 50 + n 2 – 9n
n 2 – 9n – 50 – 12n + 48 = 0
n 2 – 21n + 98 = 0
n 2 – 14n – 7n + 98 = 0
n(n – 14) – 7(n – 14) = 0
(n – 7) (n – 14) = 0
n – 7 = 0 or n – 14 = 0
n = 7 or n = 14
(2) 81
Explaination:
Given 1 + 3 + 5 + 7 + ………… + 17
This is an A.P. with first term a = 1
Common difference d = 2
Last term t n = 17
a + (n – 1)d = 17
1 + (n – 1)2 = 17
(n – 1)2 = 17 – 1
(n – 1)2 = 16
n – 1 = \(\frac{16}{2}\) = 8
n = 9