# GSEB Solutions Class 11 Statistics Chapter 6 Permutations, Combinations and Binomial Expansion Ex 6.3

Gujarat Board Statistics Class 11 GSEB Solutions Chapter 6 Permutations, Combinations and Binomial Expansion Ex 6.3 Textbook Exercise Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Statistics Chapter 6 Permutations, Combinations and Binomial Expansion Ex 6.3

Question 1.
Obtain the expansion of the following binomial expressions:
(1) (3a + 4b)3
(3a + 4b)3 = 3C0 (3a)3 (4b)0 + 3C1 (3a)2 (4b)1 + 3C2 (3a) (4b)2 + 3C3 (3a)0 (4b)3
= 1 . 27a3. 1 + 3 . 9a2 . 4b + 3 . 3a . 16b2 + 1 . 1 . 64b3
= 27a3 + 108a2b + 144b2 + 64b3

(2) (1 + x)7

(3) $$\left(\frac{3}{x}-\frac{4 x}{3}\right)^{4}$$

(4) $$\left(\frac{\sqrt{x}}{3}+\frac{3}{\sqrt{x}}\right)^{6}$$

(5) $$\left(\frac{a}{2}-\frac{b}{3}\right)^{5}$$

Question 2.
Obtain the values using binomial expansion:
(1) (âˆš5 +1)5 – (âˆš5 – 1)5

(2)(âˆš2 + 1)6 + (âˆš2 – 1)6

(3) (âˆš5 + âˆš3)4 + (âˆš5 – âˆš3)4

Question 3.
Expand (1 + x)5 and verify by putting x = 1 on both sides.
(1 + x)5 = 5C0 . 1 . x0 + 5C1 . 1 . x1 + 5C2 1 x2 + 5C3 . 1 . x3 + 5C4 1 x4 + 5C5 . 1 . x5
= 1+ 5x + 10x2 + 10x3 + 5x4 + x5
LHS = (1 + x)5
Putting x = 1,
LHS = (1 + 1)5 = (2)5 = 32
RHS = 1 + 5x + 10x2 + 10x3 + 5x4 + 1
Putting x = 1.
RHS= 1 +5(1)+ 10(1)2+ 10(1)3 + 5(1)4 + 15
= 1 + 5 + 10 + 10 + 5 + 1 = 32
Hence, LHS = RHS

Question 4.
Expand (1 + a)6 and verify by putting a = 2 on both sides.