# Question 9 Exercise 7.2

Solutions of Question 9 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.

Q9 Find numericalty greatest term in the expansion $(x-y)="$ when $x=12$ and $y-4$. Solution: When we put $x=12$, then the give becounes \begin{aligned} & \left(x \quad y=20(12-y)^{20}\right. \\ & =12^{2 n}\left(\begin{array}{ll} 1 & \frac{y}{12} \end{array}\right)^{31} \end{aligned}

Now we theck $\frac{(n+1) \cdot x}{1+|x|}$ for $\left(\frac{1}{12}\right)^2 \cdot$ Here $n=20$ and $\left.\left|x^{\prime}=\right|-\frac{y}{12} \right\rvert\,=\frac{1}{3}$ for $y=4$ $\Rightarrow \frac{(n+1) ; 4}{1+|x|}=\frac{(20+1) \cdot \frac{1}{3}}{1+\frac{1}{3}}$ $$7 \cdot \frac{3}{4}=\frac{21}{4}=5+0.25=p+F \text {. }$$

Thus $5+1=6^{t t}$ of the $12^{201}\left(1-\frac{y}{12}\right)^{20}$ is greatest i.c. $$T_6=12^{20} \cdot \frac{2(0 !}{(20-5) ! 5 !} \cdot\left(-\frac{y}{12}\right)^5$$

For $y=4$, it becomes \begin{aligned} & T_6=12^{20} \frac{20 !}{15 ! 5 !} \cdot\left(-\frac{1}{3}\right)^2 \\ & \left.=T_6-12^{20} \cdot 1-1\right)^5 \cdot \frac{20 !}{15 ! 5 !} \cdot \frac{1}{3^5} \\ & =T_6=-12^{20} \cdot 15504 \cdot \frac{1}{243} \\ & =-24460335 \div 54230.438655932 \mathrm{~K} . \end{aligned}

Ilenec this term $T_\kappa$ is numerically greates writlen abeve