7.Binomial Theorem
hard

If the coefficients of $x$ and $x^{2}$ in the expansion of $(1+x)^{p}(1-x)^{q}, p, q \leq 15$, are $-3$ and $-5$ respectively, then the coefficient of $x ^{3}$ is equal to $............$

A

$22$

B

$23$

C

$52$

D

$53$

(JEE MAIN-2022)

Solution

Since coefficient of $x$ is $-3$

${ }^{p} C _{1}-{ }^{9} C _{1}=-3$

$p – q =-3$

$\text { Comparing coefficients of } x ^{2}$

${ }^{9} C _{1}{ }^{9} C _{1}+{ }^{ p } C _{2}+{ }^{9} C _{2}=-5$

$- pq +\frac{ p ( p -1)}{2}+\frac{ q ( q -1)}{2}=-5$

Solving $(1)$ and $(2)$

$p=8, q=11$

Coefficient of $x ^{3}$ is

$-{ }^{4} C_{3}+{ }^{P} C_{3}+{ }^{P} C_{1}^{9} C_{2}-{ }^{P} C_{2}^{9} C_{1}$

$=-{ }^{11} C_{3}+{ }^{8} C_{3}+{ }^{8} C_{1}^{11} C_{2}-{ }^{8} C_{2}^{11} C_{1}$

$=23$

Standard 11
Mathematics

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