 # Permutation and Combinations NCERT Solution of Class 11 maths Chapter 7 Exercise 7.3

Question 1
How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?
3-digit numbers are to formed using the digits 1 to 9.
Important point to Note
a) The order of the digits matters
b) The repetition is not allowed
Therefore, this is a permutation of 9 different digits taken 3 at a time.
Therefore, required number of 3-digit numbers is given by
=9P3
=9! / (9-3)!
=9! /6!  =9×8×7=504
Question 2
How many 4-digit numbers are there with no digit repeated?
Important point to Note
a) The order of the digits matters
b) The repetition is not allowed
c) Thousand places cannot be zero
Since Thousand place cannot be zero, the thousands place of the 4-digit number is to be filled with any of the digits from 1 to 9.
Therefore, the number of ways in which thousands place can be filled is 9.
Now The hundreds, tens, and units place can be filled by any of the digits from 0 to 9. However, Since the digits cannot be repeated in the 4-digit numbers and thousands place is already occupied with a digit. The hundreds, tens, and units place is to be filled by the remaining 9 digits.
Therefore, this is a permutation of 9 different digits taken 3 at a time
=9P3
=9! / (9-3)!
=9! /6!  =9×8×7=504
Hence, by multiplication principle, the required number of 4-digit numbers
= 9 × 504
= 4536
Question 3
How many 3-digits even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?
Important point to Note
a) The order of the digits matters
b) The repetition is not allowed
c) Unit digit should be 2,4,6 as even numbers are required
Since the number is even, unit digits can be filled in 3 ways by any of the digits, 2, 4, or 6.
Now The hundreds and tens places can be filled by any of the digits from 6 numbers However, Since the digits cannot be repeated in the 3-digit numbers and unit place is already occupied with a digit. The hundreds and tens place is to be filled by the remaining 5 digits.
Therefore, this is a permutation of 5 different digits taken 2 at a time
=5P2
=5! / (5-2)!
=5! /3!  =5×4=20
Hence, by multiplication principle, the required number of 3-digit numbers
= 3 × 20
= 60

Question 4
Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?
4-digit numbers are to be formed using the digits, 1, 2, 3, 4, and 5.
Therefore, this is a permutation of 5 different digits taken 4 at a time
Required number of 4-digit numbers
=5P4
=5! / (5-4)!
=5! /1!  =5×4×3×2=120
Now Among the 4-digit numbers formed by using the digits, 1, 2, 3, 4, 5, even numbers end with either 2 or 4.
Important point to Note
a) The order of the digits matters
b) The repetition is not allowed
c) Unit digit should be 2,4 as even numbers are required
Now the Thousand, hundreds and tens places can be filled by any of the digits from 5 numbers However, Since the digits cannot be repeated in the 4-digit numbers and unit place is already occupied with a digit. The Thousand, hundreds and tens place is to be filled by the remaining 4 digits.
Therefore, this is a permutation of 4 different digits taken 3 at a time
=4P3
=4! / (4-3)!
=4! /1!  =4×3×2=24
Thus, by multiplication principle, the required number of even numbers
= 24 × 2 = 48

Question 5
From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person cannot hold more than one position?
From a committee of 8 persons, a chairman and a vice chairman are to be chosen in such a way that one person cannot hold more than one position.
Here, the number of ways of choosing a chairman and a vice chairman is the permutation of 8 different objects taken 2 at a time.
Hence, required number of ways
=8P2
=8! / (8-2)!
=8! /6!  =8×7=56

Question 6
Find n if
n-1P3:nP4 = 1 :9
n-1P3:nP4 = 1 :9
(n-1)! / (n-4)! : n ! : (n-4)! =1 :9
(n-1)! : n! = 1:9
1 : n= 1:  9
n=9
Question 7
Find r if
i)5 Pr =2 6 Pr-1
ii) 5 Pr = 6 Pr-1
(i) 5 Pr =2 6 Pr-1
5! / (5-r)! = 2 6! / (7-r)  !
1 / (5-r)! = 12 / (7-r) (6-r)(5-r)!
1= 12/ (7-r)(6-r)
(7-r) (6-r) =12
r2 -13 r +30 =0
(r-3) (r-10) =0
It is known that,
0 ≤ r ≤ 5
Hence, r ≠ 10
r = 3
(ii)
5 Pr = 6 Pr-1
5! / (5-r)! =  6! / (7-r)  !
1 / (5-r)! = 6 / (7-r) (6-r)(5-r)!
1= 6/ (7-r)(6-r)
(7-r) (6-r) =6
r2 -13 r +36 =0
(r-4) (r-9) =0
It is known that,
0 ≤ r ≤ 5
Hence, r ≠ 9
r = 4
Question 8
How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?
There are 8 different letters in the word EQUATION.
Therefore, the number of words that can be formed using all the letters of the word
EQUATION, using each letter exactly once, is the number of permutations of 8 different objects taken 8 at a time
=8P8
=8! / (8-8)!
=8! /0!  =8×7×6×5×4×3×2=40320

Question 9
How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if
(i) 4 letters are used at a time,
(ii) all letters are used at a time,
(iii) all letters are used but first letter is a vowel?
We have 6 different letters in the word MONDAY.
(i) Number of 4-letter words that can be formed from the letters of the word MONDAY, without repetition of letters, is the number of permutations of 6 different objects taken 4 at a time
=6P4
=6! / (6-4)!
=6! /2!  =6×5×4×3=360
(ii) Number of words that can be formed by using all the letters of the word MONDAY at a time is the number of permutations of 6 different objects taken 6 at a time,
=6P6
=6! / (6-6)!
=6! /0!  =6×5×4×3×2=720
(iii) Important point to Note
a) The order of the word matters
b) The repetition is not allowed
c) first letter is a vowel
In the given word, there are 2 different vowels, which have to occupy the rightmost place of the words formed. Hence This can be done only in 2 ways.
Now Since the letters cannot be repeated and the rightmost place is already occupied with a letter (which is a vowel), the remaining five places are to be filled by the remaining 5 letters. Therefore, this is a permutation of 5 different objects  taken 5 at a time
=5P5
=5! / (5-5)!
=5! /0!  =5×4×3×2=120
Hence, in this case, required number of words that can be formed is
5! × 2 = 120 × 2 = 240

Question 10
In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?
In the given word MISSISSIPPI, I appear 4 times, S appears 4 times, P appears 2 times, and M appears just once.
hence, number of distinct permutations of the letters in the given word
=11! / 4! 4! 2! = 34650
There are 4 Is in the given word. When they occur together, they are treated as a single object for the time being. This single object together with the remaining 7 objects will account for 8 objects.
These 8 objects in which there are 4 Ss and 2 Ps can be arranged in ways
=8! / 4!  2!
=840 ways.
So, Number of arrangements where all Is occurred together = 840
Therefore, number of distinct permutations of the letters in MISSISSIPPI in which four Is do not come together = 34650 – 840 = 33810

Question 11
In how many ways can the letters of the word PERMUTATIONS be arranged if the
(ii)vowels are all together,
(iii) there are always 4 letters between P and S?
Given the word PERMUTATIONS
a) PERMUTATIONS, there are 2 Ts
b) all the other letters appear only once.
c) Total 12 letters
(i) If P and S are fixed at the extreme ends (P at the left end and S at the right end), then 10 letters are left.
Hence, in this case, required number of arrangements
=10! / 2!
=1814400
(ii) There are 5 vowels in the given word, each appearing only once.
Since they must always occur together, they can be clubbed as a single object for the time being. This single object together with the remaining 7 objects will account for 8 objects.
These 8 objects in which there are 2 Ts can be arranged in the ways
=8! / 2!
Corresponding to each of these arrangements, the 5 different vowels can be arranged in the ways
=5!
Therefore, by multiplication principle, required number of arrangements in this case
= (8! / 2!) × 5!
=2419200
(iii) The letters must be arranged in such a way that there are always 4 letters between P and S.
Therefore, in a way, the places of P and S are fixed. The remaining 10 letters can be arranged
=10! / 2!
Also, 4 letters can be placed in between P and S in 2 × 7 = 14 ways.
Hence, by the principle of multiplication, number of arrangements required in this case = (10! / 2!) ×14 = 25401600.