# Question 3 & 4 Review Exercise 7

Solutions of Question 3 & 4 of Review Exercise 7 of Unit 07: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Q3 What is the coefficient of the fourth term in the expansion of $(2 x-4 y)^7$ ? Solution: Here $n=7, a=2 x$ and $b=-4 y$. The fourth tenn of the above expansion is: \begin{aligned} & T_{3+1}=\frac{7 !}{(7-3) ! 3 !}(2 x)^{7 \cdot 3}(-4 y)^3 \\ & =\frac{7 !}{(7-3) ! 3 !} \cdot\left(2^4\right) \cdot(-4)^3 \cdot x^4 y^3 \\ & \Rightarrow T_4=-35840 x^4 y^3 . \end{aligned}

Q4 $2^7 x y^3$ is a term in the expansion of $(a x+2 y)^4$. Find $a$. Solution: Here $n=7, a^{\prime}=2 x$ and $b=-4 y$. Let $T_{r+1}$ be the term $2^7 x y^3$ of the above expansion, that is: \begin{aligned} & T_{r+1}=\frac{4 !}{(4-r)+\infty !}(a x)^{4-r}(2 y)^r \\ & =\frac{4 !}{(4-r) ! r !} 2^r a^4 x^{4-r} y^r \end{aligned}

But $T_{i+1}=2^7 x y^3$ is possible only if $$x^{4 \prime \prime} y^{\prime}=x y^3 \Rightarrow r=3 \text {. }$$

Thus the term is: \begin{aligned} & T 3+1=\frac{4 !}{(4-3) ! 3 !} 2^3 a^4 x^{4-3} y^3 . \\ & =2^7 x y^3 \\ & \Rightarrow 4 \cdot 2^3 \cdot a^4=2^7 \\ & \Rightarrow a^4=\frac{2^7}{2^3 \cdot 2^2}=2^2 \\ & \Rightarrow a^2=\sqrt{2^2}=2 \\ & \Rightarrow a= \pm \sqrt{2} . \end{aligned}