The coefficient of `x^(10)` in the expansion of `(1+x^2-x^3)^8` is `476` b. `496` c. `506` d. `528`

Options

A. 476
B. 496
C. 506
D. 528

Answer

A

Explanation

We rewrite the given expression as `[1+x^(2)(1-x)]^(8)` and expand by using the binomial theorem. We have,
`[1+x^(2)(1-x)]^(8)=.^(8)C_(0)+.^(8)C_(1)x^(2)(1-x)+.^(8)C_(2)x^(4)(1-x)^(2)+.^(8)C_(3)x^(6)(1-x)^(3)+.^(8)C_(4)x^(8)(1-x)^(4)+.^(8)C_(5)x^(10)(1-x)^(5)+"....."`
The two terms which contain `x^(10)` are `.^(8)C_(4) x^(8)(1-x)^(4)` and `.^(8)C_(5)x^(10)(1-x)^(5)`
Thus, the coefficient of `x^(10)` in the given expression is given by `.^(8)C_(4)` [coefficient of `x^(2)` in the expansion of `(1-x)^(4)`]`+.^(8)C_(5)`.
`= .^(8)C_(4)(6)+.^(8)C_(5)=(8!)/(4!4!) (6) + (8!)/(3!5!)`
`= (70)(6) + 56 = 476`