NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Exercise 7.3

Chapter 7 Permutations and Combinations Exercise 7.3 Class 11 Maths NCERT Solutions is given here that can be used to score better marks in the examinations. NCERT Solutions for Class 11 Maths prepared by Studyrankers are helpful in revising the syllabus as it updated as per the latest guidelines of CBSE. These can be used to prepare your own answers and solve hard problems easily.

NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Exercise 7.3

1. How many 3-digit number can be formed by using the digits 1 to 9 if no digit is repeated?

Answer

3 digit number are to be form with digits 1 to 9. 
This can be done in 9 × 8 × 7 = 504 ways.

2. How many 4-digit numbers are there with no digit repeated?

Answer

The digits be 0 to 9 
4 digit number 10P4
This includes those number which have 0 in the beginning 
3-digit numbers out of 9 digits 1 to 9 are 9P3
∴ 4 digit numbers which do not have zero in the beginning = 10P4 – 9P3
= 10 × 9 × 8 × 7 – 9 × 8 × 7 = 9 × 9 × 8 × 7 = 4536

3. How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no. digit is repeated.

Answer

One of the digit 2, 4 and 6 will come at unit place. 
∴ The unit place can be filled in 3P1 ways. 
Now, we have 5 digits and 2 places are to be filled up 
This can be done in 5Pways. 
No. of 3-digit even numbers are 3P1 × 5P2
= 3 × 20 = 60

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?

Answer

(i) Out of 5 digits 4-digit numbers are to be formed. Such numbers are 
5P4 = 5 × 4 × 3 × 2 = 120 
(ii) When 2 is at units place then remaining three place are filled in 4P3 ways 
= 4 × 3 × 2 = 24 
When 4 is at unit’s place, then 4-digit numbers are again = 24 
∴ Even 4- digit numbers = 2 × 24 = 48

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?

Answer

Out of 8 persons chairman can be chosen in 8 ways. After selecting a chairman, we have to choose a vice-chairman out of 7 persons. This can be done in 7 ways.
Out of 8 persons, a chairman and vice chairman can be choosen in 8 × 7 = 56 ways.

6. Find n if n – 1P3nP4 = 1: 9

Answer

n-1P3/nP4 = 1/9 => n-1P3/n.n-1P3 = 1/9
=> 1/n = 1/9 or n = 9

7. Find r if (i) 5Pr = 2 6Pr – 1 (ii) 5Pr = 6Pr – 1 

Answer

(i) We have,
5Pr = 2 6Pr – 1
or 5!/(5 – r)! = 2[6!/(6 – r + 1)!]
or 5!/(5 – r)! = 2[(6×5)!/(7 – r)!]
or 1/(5 – r)! = 12/((7 – r)(6 – r)(5 – r)!)
or (7 – r) (6 – r) = 12
or 42 – 7r – 6 r + r 2 = 12
or r 2 – 13 r + 30 = 0
or r2 – 10 r – 3r + 30 = 0
or r (r – 10) – 3 (r – 10) = 0
or (r – 10) (r – 3) = 0
or r = 10 or r = 3
Hence, r = 3
[r = 10 => 5P10 which is meaningless]
(ii) (ii) We have,
5Pr = 6Pr – 1
or 5!/(5 – r)! = 2[6!/[6-(r – 1)]!]
or 5!/(5 – r)! = (6 × 5!)/(6 – r + 1)!
or 5!/(5 – r)! = (6 × 5!)/(7 – r )!
or 5!/(5 – r)! = (6 × 5!)/((7 – r)(6 – r)(5 – r)!
or (7 – r) (6 – r) = 6
or r 2 – 13 r + 36 = 0
or r = 4, 9
or r = 4
[r = 9 => 5Pr which is meaningless]

8. How many words with or without meaning, can be formed using all the letters of the word EQUATION. Using each letter exactly once?

Answer

Number of letters in EQUATION = 8 
Number of letters to be taken at a time = 8 
If P is the number of words thus formed, then 
P = 8P8 = 8!/(8 - 8)! - = 8! 
= 8 × 7 × 6 × 5 × 4 × 3 × 2 
= 40320

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?

Answer

(i) MONDAY have 6 letters 
4 letters are taken at a time 
Number of words = 6P
= 6 × 5 × 4 × 3 = 360 
(ii) All the letters of word MONDAY are taken at a time 
Number of words = 6! 
= 6 ×5 × 4 × 3 × 2 × 1 = 720 
(iii) Let the words begin with A 
Number of words formed with 5 letters 
= 5! = 120 
Similarly, when words begin with O number of such words = 120 
Numbers of words begin with vowel 
= 2 × 120 = 240

10. In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together? 

Answer

In given word there are 4I, 4S, 2P and 1M.
Total number of permutations = 11! /(4!4!2!)
If take 4 I’s as one letter then total letters become
= 11 – 4 +1 = 8
If P is the permutations when 4 I’s are not together, then P = 11! /(4!4!2!) - 8!/ (4!2!)
= (11 × 10 × 9 ×8 ×7 ×6 ×5 ×4 ×3 ×2 ×1)/(4 × 3 ×2 ×1 × 2 × 1 × 4!) - (8 × 7 ×6 ×5 × 4!)/(2 × 1 × 4!)
= 34650 – 840 = 33810

11. In how many ways can the letters of the word PERMUTATIONS be arranged if the
(i) words start with P and end with S
(ii) vowels are all together,
(iii) there are always 4 letters between P and S?

Answer

(i) Letters between P and S are
ERMUTATION.
These 10 letters having T two times,
these letters can be arranged in 10! /2! ways
= 1814400 ways

(ii) There are 12 letters in the word PERMUTATIONS. which have T two times.
Now the vowels a, e, i, o, u are together
Let it be considered in one block.
The letters of vowels can be arranged in 5! ways
Thus, there are 7 letters and 1 block of vowel with T two times
∴ Number of arrangements = (8!/5!) × 2!
= 2419200

(iii) There are 12 letters to be arranged in 12 place [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] there are 12 letters are to be filled in 12 place. P may be filled up at place No. 1, 2, 3, 4, 5, 6, 7 and consequently S may be filled up at place no. 6, 7, 8, 9, 10, 11, 12. leaving four places between. Now P and S may be filled up in 7 ways.
Similarly, S and P may be filled up in 7 ways
Thus, P and S or S and P can be filled up in 7 + 7 = 14 ways
Now, the remaining 10 places can be filled by
=10!/ 2! = ways
∴ No of ways in which 4 letters occur between P and S
= (10!/2!) ×14 = (3628800/2) × 14 = 25401600
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