Q. Find the coefficient of $x^{5}$ in $\left(1+x+x^{2}\right)^{8}$.
AP EAMCETAP EAMCET 2020
Solution:
Given, Expansion is $\left(1+x+x^{2}\right)^{8}$
Coefficient of $x^{5}$ in above Expansion is
$\sum \frac{8 !}{n_{1} ! n_{2} ! n_{3} !}(1)^{n_{1}} \cdot(x)^{n_{2}}\left(x^{2}\right)^{n_{3}}$
$\therefore \left(n_{1}+n_{2}+n_{3}\right)=8$
and $n_{2}+2 n_{3}=5$
$n_1$
$n_2$
$n_3$
$5$
$3$
$0$
$5$
$1$
$2$
$4$
$3$
$1$
Coefficient of $x^{5}=\frac{8 !}{5 ! 3 ! 0 !}+\frac{8 !}{5 ! 1 ! 2 !}+\frac{8 !}{4 ! 3 ! 1 !}$
$=56+168+280=504$
| $n_1$ | $n_2$ | $n_3$ |
| $5$ | $3$ | $0$ |
| $5$ | $1$ | $2$ |
| $4$ | $3$ | $1$ |