# Question 4 Exercise 7.2

Solutions of Question 4 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.

Find the coefticient of $x^{23}$ in $(x^2-x)^{20}$

In the above expansion $n=20, \quad a=x^2$ and $b=-x$.

Let $T_{r, 1}$ be the term containing $x^{23}$ that is: \begin{align}T_{r-1}&=\dfrac{20 !}{(20-r) ! r !}(x^2)^{20 r}(-x)^r \\ & =\dfrac{20 !}{(20-r) ! r !}(-1)^r \cdot x^{40-2 r+r} \\ & =\dfrac{20 !}{(20-r) ! r !}(-1)^r x^{40-r}\end{align} But $T_{r-1}$ containing $x^{33}$ is possibic only if \begin{align}x^{40 \cdot r}&=x^{23}\\ \Rightarrow 40-r&=23\\ \Rightarrow r&=40-23=17\end{align} Putting $r=17$. in $T_{r+1}$ we get \begin{align}T_{17-1}&=\dfrac{20 !}{(20 \cdot 17) ! 17 !}(-1)^{17} x^{40-17} \\ \Rightarrow T_{18}&=-1140 x^{23}\end{align} Hence the coefficient of $x^{23}$ in the expansion of $(x^2-x^{20})$ is $-1140$

Find the coefticient of $\dfrac{1}{x^4}$ in $(2-\dfrac{1}{x})^x$

In the above expansion $n=8, \quad a=2$ and $b=-\dfrac{1}{x}$.

Let $T_{r + 1}$ be the term containing $x^{23}$ that is:

\begin{align}T_{r+1}&=\dfrac{8 !}{(8-r) ! r !}(2)^{8-r}(-\dfrac{1}{x})^r\\ &=\dfrac{8 !}{(8-r) ! r!}2^{8-r}(-1)^r(\dfrac{1}{x})^r\end{align} But $T_{r+1}$ conlaining $x^4$ is possible only if \begin{align}\dfrac{1}{x^r}&=\dfrac{1}{x^4}\\ \Rightarrow 4=r\end{align} Putting $r=4$ in $T_{r+ 1}$ we get \begin{align}T_{4+1}&=\dfrac{8 !}{(8-4)!4!}2^4(-1)^4(\dfrac{1}{x^4})\\ \Rightarrow T_5&=^8C_4\, 2^4 x^{-4} \end{align} Hence the coefficient of $\dfrac{1}{x^{-1}}$ in she expansion of $(2-\dfrac{9}{x})$, is $^8C_4\, 2^4$

Find the coefticient of $a^6 b^3$ in $(2 a-\dfrac{b}{3})^9$

In the above expansion $n=9 . \quad a=2a$ and $b=-\dfrac{b}{3}$.

Let \begin{align}T_{r+1}&= \dfrac{9 !}{9-r)!r !}(2 a)^{9-r}(-\dfrac{b}{3})^r\\ T_{R+1}&=\dfrac{9 !}{(9-r) ! r !})2^{9-r} \cdot(\dfrac{1}{3^r}) a^{9-r} \cdot b^r\end{align} But the term $T_{r+1}$ tontaining $a^6 b^3$ is possible only if \begin{align}a^{9-r}b^r&=a^6 b^3\\ \Rightarrow 9 -r=6 \text{or} r=3\end{align} Putting $r=3$ in the above, we get \begin{align}T_{3+1}&=\dfrac{9 !}{(9-3) ! 3 !} 2^{9-3} \cdot(-\dfrac{1}{3})^3 a^{9-3}b^3 \\ T_4&=-^9C_3 2^6 \frac{1}{3^3} a^6 b^3 \\ \Rightarrow T_4=-^9C_3 \dfrac{2^6}{27} a^6 b^3 \end{align} Henee the coeficient of $a^6 b^3$ in the expansion of $(2 a-\dfrac{b}{3})$ is $-^9C_3 \dfrac{2^6}{27}$