Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Permutations and Combinations Ex 6.2 Questions and Answers.

## Maharashtra State Board 11th Commerce Maths Solutions Chapter 6 Permutations and Combinations Ex 6.2

Question 1.

Evaluate:

(i) 8!

Solution:

8!

= 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1

= 40320

(ii) 6!

Solution:

6!

= 6 × 5 × 4 × 3 × 2 × 1

= 720

(iii) 8! – 6!

Solution:

8! – 6!

= 8 × 7 × 6! – 6!

= 6! (8 × 7 – 1)

= 6! (56 – 1)

= 6 × 5 × 4 × 3 × 2 × 1 × 55

= 39,600

(iv) (8 – 6)!

Solution:

(8 – 6)!

= 2!

= 2 × 1

= 2

Question 2.

Compute:

(i) \(\frac{12 !}{6 !}\)

Solution:

\(\frac{12 !}{6 !}\)

= \(\frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 !}{6 !}\)

= 12 × 11 × 10 × 9 × 8 × 7

= 665280

(ii) \(\left(\frac{12}{6}\right) !\)

Solution:

\(\left(\frac{12}{6}\right) !\)

= 2!

= 2 × 1

= 2

(iii) (3 × 2)!

Solution:

(3 × 2)!

= 6!

= 6 × 5 × 4 × 3 × 2 × 1

= 720

(iv) 3! × 2!

Solution:

3! × 2!

= 3 × 2 × 1 × 2 × 1

= 12

Question 3.

Compute:

(i) \(\frac{9 !}{3 ! 6 !}\)

Solution:

(ii) \(\frac{6 !-4 !}{4 !}\)

Solution:

(iii) \(\frac{8 !}{6 !-4 !}\)

Solution:

(iv) \(\frac{8 !}{(6-4) !}\)

Solution:

Question 4.

Write in terms of factorials

(i) 5 × 6 × 7 × 8 × 9 × 10

Solution:

5 × 6 × 7 × 8 × 9 × 10

= 10 × 9 × 8 × 7 × 6 × 5

Multiplying and dividing by 4!, we get

(ii) 3 × 6 × 9 × 12 × 15

Solution:

3 × 6 × 9 × 12 × 15

= 3 × (3 × 2) × (3 × 3) × (3 × 4) × (3 × 5)

= (3^{5}) (5 × 4 × 3 × 2 × 1)

= 3^{5} (5!)

(iii) 6 × 7 × 8 × 9

Solution:

6 × 7 × 8 × 9

= 9 × 8 × 7 × 6

Multiplying and dividing by 5!, we get

(iv) 5 × 10 × 15 × 20 × 25

Solution:

5 × 10 × 15 × 20 × 25

= (5 × 1) × (5 × 2) × (5 × 3) × (5 × 4) × (5 × 5)

= (5^{5}) (5 × 4 × 3 × 2 × 1)

= (5^{5}) (5!)

Question 5.

Evaluate: \(\frac{n !}{r !(n-r) !}\) for

(i) n = 8, r = 6

Solution:

n = 8, r = 6

(ii) n = 12, r = 12

Solution:

n = 12, r = 12

Question 6.

Find n, if

(i) \(\frac{n}{8 !}=\frac{3}{6 !}+\frac{1}{4 !}\)

Solution:

(ii) \(\frac{n}{6 !}=\frac{4}{8 !}+\frac{3}{6 !}\)

Solution:

(iii) \(\frac{1}{n !}=\frac{1}{4 !}-\frac{4}{5 !}\)

Solution:

Question 7.

Find n, if

(i) (n + 1)! = 42 × (n – 1)!

Solution:

(n + 1)! = 42(n – 1)!

∴ (n + 1) n (n – 1)! = 42(n – 1)!

∴ n^{2} + n = 42

∴ n(n + 1) = 6 × 7

Comparing on both sides, we get

∴ n = 6

(ii) (n + 3)! = 110 × (n + 1)!

Solution:

(n + 3)! = 110 × (n + 1)!

∴ (n + 3) (n + 2) (n + 1)! = 110 (n + 1)!

∴ (n + 3) (n + 2) = (11) (10)

Comparing on both sides, we get

n + 3 = 11

∴ n = 8

Question 8.

Find n, if:

(i) \(\frac{n !}{3 !(n-3) !}: \frac{n !}{5 !(n-5) !}=5: 3\)

Solution:

∴ 12 = (n – 3)(n – 4)

∴ (n – 3)(n – 4) = 4 × 3

Comparing on both sides, we get

n – 3 = 4

∴ n = 7

(ii) \(\frac{n !}{3 !(n-5) !}: \frac{n !}{5 !(n-7) !}=10: 3\)

Solution:

∴ (n – 5) (n – 6) = 3 × 2

Comparing on both sides, we get

n – 5 = 3

∴ n = 8

Question 9.

Find n, if:

(i) \(\frac{(17-n) !}{(14-n) !}\) = 5!

Solution:

\(\frac{(17-n) !}{(14-n) !}\) = 5!

∴ \(\frac{(17-n)(16-n)(15-n)(14-n) !}{(14-n) !}\) = 5 × 4 × 3 × 2 × 1

∴ (17 – n) (16 – n) (15 – n) = 6 × 5 × 4

Comparing on both sides, we get

17 – n = 6

∴ n = 11

(ii) \(\frac{(15-n) !}{(13-n) !}\) = 12

Solution:

\(\frac{(15-n) !}{(13-n) !}\) = 12

∴ \(\frac{(15-\mathrm{n})(14-\mathrm{n})(13-\mathrm{n}) !}{(13-\mathrm{n}) !}\) = 12

∴ (15 – n) (14 – n) = 4 × 3

Comparing on both sides, we get

15 – n = 4

∴ n = 11

Question 10.

Find n if \(\frac{(2 n) !}{7 !(2 n-7) !}: \frac{n !}{4 !(n-4) !}\) = 24 : 1

Solution:

∴ (2n – 1) (2n – 3) (2n – 5) = 9 × 7 × 5

Comparing on both sides, we get

2n – 1 = 9

∴ n = 5

Question 11.

Show that \(\frac{n !}{r !(n-r) !}+\frac{n !}{(r-1) !(n-r+1) !}=\frac{(n+1) !}{r !(n-r+1)}\)

Solution:

Question 12.

Show that \(\frac{9 !}{3 ! 6 !}+\frac{9 !}{4 ! 5 !}=\frac{10 !}{4 ! 6 !}\)

Solution:

Question 13.

Find the value of:

(i) \(\frac{8 !+5(4 !)}{4 !-12}\)

Solution:

(ii) \(\frac{5(26 !)+(27 !)}{4(27 !)-8(26 !)}\)

Solution:

Question 14.

Show that

\(\frac{(2 n) !}{n !}\) = 2^{n} (2n – 1) (2n – 3)…5.3.1

Solution: