## 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.

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 Ã— 8 Ã— 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

^{10}P_{4}
This includes those number which have 0 in the beginning

3-digit numbers out of 9 digits 1 to 9 are

^{9}P_{3}
âˆ´ 4 digit numbers which do not have zero in the beginning =

^{10}P_{4}â€“^{9}P_{3}
= 10 Ã— 9 Ã— 8 Ã— 7 â€“ 9 Ã— 8 Ã— 7 = 9 Ã— 9 Ã— 8 Ã— 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.

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.

âˆ´ The unit place can be filled in

^{3}P_{1}ways.
Now, we have 5 digits and 2 places are to be filled up

This can be done in

^{5}P_{2 }ways.
No. of 3-digit even numbers are

^{3}P_{1}Ã—^{5}P_{2}
= 3 Ã— 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?

or 5!/(5 â€“ r)! = 2[6!/(6 â€“ r + 1)!]

or 5!/(5 â€“ r)! = 2[(6Ã—5)!/(7 â€“ r)!]

or 1/(5 â€“ r)! = 12/((7 â€“ r)(6 â€“ r)(5 â€“ r)!)

or (7 â€“ r) (6 â€“ r) = 12

or 42 â€“ 7r â€“ 6 r + r

or r

or r

or r (r â€“ 10) â€“ 3 (r â€“ 10) = 0

or (r â€“ 10) (r â€“ 3) = 0

or r = 10 or r = 3

Hence, r = 3

[r = 10 =>

(ii)

or 5!/(5 â€“ r)! = 2[6!/[6-(r â€“ 1)]!]

or 5!/(5 â€“ r)! = (6 Ã— 5!)/(6 â€“ r + 1)!

or 5!/(5 â€“ r)! = (6 Ã— 5!)/(7 â€“ r )!

or 5!/(5 â€“ r)! = (6 Ã— 5!)/((7 â€“ r)(6 â€“ r)(5 â€“ r)!

or (7 â€“ r) (6 â€“ r) = 6

or r

or r = 4, 9

or r = 4

[r = 9 =>

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

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

^{5}P

_{4}= 5 Ã— 4 Ã— 3 Ã— 2 = 120

**(ii)**When 2 is at units place then remaining three place are filled in

^{4}P

_{3}ways

= 4 Ã— 3 Ã— 2 = 24

When 4 is at unitâ€™s place, then 4-digit numbers are again = 24

âˆ´ Even 4- digit numbers = 2 Ã— 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?

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 Ã— 7 = 56 ways.

6. Find n if

=> 1/n = 1/9 or n = 9

7. Find r if (i)

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 Ã— 7 = 56 ways.

6. Find n if

^{n â€“ 1}P_{3}:^{n}P_{4}= 1: 9**Answer**^{n-1}P_{3}/^{n}P_{4}= 1/9 =>^{n-1}P_{3}/n.^{n-1}P_{3}= 1/9=> 1/n = 1/9 or n = 9

7. Find r if (i)

^{5}P_{r}= 2^{6}P_{r â€“ 1}(ii)^{5}P_{r}=^{6}P_{r â€“ 1 }**Answer****(i)**We have,^{5}P_{r}= 2^{6}P_{r â€“ 1}or 5!/(5 â€“ r)! = 2[6!/(6 â€“ r + 1)!]

or 5!/(5 â€“ r)! = 2[(6Ã—5)!/(7 â€“ r)!]

or 1/(5 â€“ r)! = 12/((7 â€“ r)(6 â€“ r)(5 â€“ r)!)

or (7 â€“ r) (6 â€“ r) = 12

or 42 â€“ 7r â€“ 6 r + r

^{2}= 12or r

^{ 2}â€“ 13 r + 30 = 0or r

^{2}â€“ 10 r â€“ 3r + 30 = 0or r (r â€“ 10) â€“ 3 (r â€“ 10) = 0

or (r â€“ 10) (r â€“ 3) = 0

or r = 10 or r = 3

Hence, r = 3

[r = 10 =>

^{5}P_{10}which is meaningless](ii)

**(ii)**We have,^{5}P_{r}=^{6}P_{r â€“ 1}or 5!/(5 â€“ r)! = 2[6!/[6-(r â€“ 1)]!]

or 5!/(5 â€“ r)! = (6 Ã— 5!)/(6 â€“ r + 1)!

or 5!/(5 â€“ r)! = (6 Ã— 5!)/(7 â€“ r )!

or 5!/(5 â€“ r)! = (6 Ã— 5!)/((7 â€“ r)(6 â€“ r)(5 â€“ r)!

or (7 â€“ r) (6 â€“ r) = 6

or r

^{ 2}â€“ 13 r + 36 = 0or r = 4, 9

or r = 4

[r = 9 =>

^{5}P_{r}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 =

^{8}P_{8}= 8!/(8 - 8)! - = 8!
= 8 Ã— 7 Ã— 6 Ã— 5 Ã— 4 Ã— 3 Ã— 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?

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 =

^{6}P_{4 }
= 6 Ã— 5 Ã— 4 Ã— 3 = 360

**(ii)**All the letters of word MONDAY are taken at a time

Number of words = 6!

= 6 Ã—5 Ã— 4 Ã— 3 Ã— 2 Ã— 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 Ã— 120 = 240

10. In how many of the distinct permutations of the letters in MISSISSIPPI do the four Iâ€™s not come together?

10. In how many of the distinct permutations of the letters in MISSISSIPPI do the four Iâ€™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 Iâ€™s as one letter then total letters become

= 11 â€“ 4 +1 = 8

If P is the permutations when 4 Iâ€™s are not together, then P = 11! /(4!4!2!) - 8!/ (4!2!)

= (11 Ã— 10 Ã— 9 Ã—8 Ã—7 Ã—6 Ã—5 Ã—4 Ã—3 Ã—2 Ã—1)/(4 Ã— 3 Ã—2 Ã—1 Ã— 2 Ã— 1 Ã— 4!) - (8 Ã— 7 Ã—6 Ã—5 Ã— 4!)/(2 Ã— 1 Ã— 4!)

= 34650 â€“ 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?

(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

âˆ´ Number of arrangements = (8!/5!) Ã— 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

âˆ´ No of ways in which 4 letters occur between P and S

= (10!/2!) Ã—14 = (3628800/2) Ã— 14 = 25401600

Total number of permutations = 11! /(4!4!2!)

If take 4 Iâ€™s as one letter then total letters become

= 11 â€“ 4 +1 = 8

If P is the permutations when 4 Iâ€™s are not together, then P = 11! /(4!4!2!) - 8!/ (4!2!)

= (11 Ã— 10 Ã— 9 Ã—8 Ã—7 Ã—6 Ã—5 Ã—4 Ã—3 Ã—2 Ã—1)/(4 Ã— 3 Ã—2 Ã—1 Ã— 2 Ã— 1 Ã— 4!) - (8 Ã— 7 Ã—6 Ã—5 Ã— 4!)/(2 Ã— 1 Ã— 4!)

= 34650 â€“ 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

âˆ´ Number of arrangements = (8!/5!) Ã— 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

âˆ´ No of ways in which 4 letters occur between P and S

= (10!/2!) Ã—14 = (3628800/2) Ã— 14 = 25401600