# Question 1 Exercise 7.2

Solutions of Question 1 of Exercise 7.2 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.

Expand by using Binomial theorem: $(x^2-\dfrac{1}{y})^4$

Using binomial theorem \begin{align}(x^2-\dfrac{1}{y})^4&=(x^2)^4+{ }^4 C_1(x^2)^3(-\dfrac{1}{y})+ \\ & { }^4 C_2(x^2)^2(-\dfrac{1}{y})^2+{ }^4 C_3(x^2)(-\dfrac{1}{y})^3 + { }^4 C_4(-\dfrac{1}{y})^4 \\ & =x^8- \dfrac{4x^6}{y}+\dfrac{6x^4}{y^2}- \dfrac{4x^2}{y^3}+\dfrac{1}{y^4} \end{align}

Expand by using Binomial theorem: $(1+x y)^7$

Using binonial theorem \begin{align} & (1+x y)^7=1+{ }^7 C_1(1)^6(x y)+{ }^7 C_2 \cdot 1^5(x y)^2 +{ }^7 C_3 1^4(x y)^3\\ &+{ }^7 C_4 1^3(x y)^4+{ }^7 C_5 1^2(x y)^5+{ }^7 C_6 1^1(x y)^6+{ }^7 C_7(x y)^7\\ &=1+7 x y+21 x^2 y^2+35 x^3 y^3+35 x^4 y^4 +21 x^5 y^5+7 x^6 y^6+x^7 y^7 \text {. } \end{align}

Expand by using Binomial theorem: $(\sqrt{y}+\dfrac{1}{\sqrt{y}})^5$

Using binomial theorem \begin{align}(\sqrt{y}+\dfrac{1}{\sqrt{y}})^5&=(\sqrt{y})^5+ { }^5 C_1(\sqrt{y})^4(\dfrac{1}{\sqrt{y}})+{ }^5 C_2(\sqrt{y})^3(\dfrac{1}{\sqrt{y}})^2+ \\ & { }^5 C_3(\sqrt{y})^2(\dfrac{1}{\sqrt{y}})^3+{ }^5 C_4(\sqrt{y})(\dfrac{1}{\sqrt{y}})^4+ { }^5 C_5(\dfrac{1}{\sqrt{y}})^5 \\ & =y^{\dfrac{5}{2}}+5 \cdot y^2 \cdot y^{-\dfrac{1}{2}}+10 \cdot y^{\dfrac{3}{2}} \cdot y^{-1}\\ &+ 10 \cdot y \cdot y^{-\dfrac{3}{2}}+5 \cdot y^{\dfrac{1}{2}} \cdot y^{-2}+y^{-\dfrac{3}{2}} \\ & =y^{\dfrac{5}{2}}y^{\dfrac{-1}{2}}y^{\dfrac{1}{2}}+5 \cdot y^{2-\dfrac{1}{2}}+10 \cdot y^{\dfrac{3}{2}-\dfrac{1}{2}-\dfrac{1}{2}}\\ &+10 \cdot y^{1-\dfrac{3}{2}} +5 \cdot y^{\dfrac{1}{2}-\dfrac{3}{2}-\dfrac{1}{2}}+y^{-\dfrac{3}{2}}\\ &=\dfrac{1}{\sqrt{y}}\left[y^{\dfrac{6}{2 }} +5 y^2+10 y^{\dfrac{2}{2}}+10 y^{1-1}+5y^{\dfrac{-2}{2}}+y^{-2}\right]\\ &=\dfrac{1}{\sqrt{y}}\left[y^{3} +5 y^2+10 y+10+ \dfrac{5}{y}+\dfrac{1}{y^2}\right]\end{align}