Coefficient problem

$\displaystyle (1+2t^2-t^3)^9=[1+t^2(2-t)]^9= \sum_{n=0} ^9 {\begin{pmatrix} 9 \\ n \end{pmatrix} t^{2n}(2-t)^n}$

There are only two cases, when $n=3$ and $n=4$ that involves $t^8$ .

$\begin{array} {llll} n = 3 & \Rightarrow \begin{pmatrix} 9 \\ 3 \end{pmatrix} t^6(2-t)^3 & \Rightarrow \begin{pmatrix} 9 \\ 3 \end{pmatrix} (-1)^2 \begin{pmatrix} 3 \\ 2 \end{pmatrix} 2 & = 504 \\ n = 4 & \Rightarrow \begin{pmatrix} 9 \\ 4 \end{pmatrix} t^8(2-t)^4 & \Rightarrow \begin{pmatrix} 9 \\ 4 \end{pmatrix} (-1)^0 \begin{pmatrix} 4 \\ 0 \end{pmatrix} 2^4 & = 2016 \end{array}$

Therefore, the coefficient of $t^8$ , $a_8 = 504+2016 = \boxed{2520}$

${ (x+y+z) }^{ n }=\sum { \frac { n! }{ p!q!r! } } { x }^{ p }{ y }^{ q }{ z }^{ r },\quad Where\quad p+q+r=n\quad \\ and\quad p,q,r\quad \in \quad N.\\ Thus\quad { (1+{ 2t }^{ 2 }-{ t }^{ 3 }) }^{ 9 }=\sum { \frac { 9! }{ p!q!r! } } { 1 }^{ p }{ (2{ t }^{ 2 }) }^{ q }{ ({ -t }^{ 3 }) }^{ r },\\ Here\quad p+q+r=9\quad .....(1)\\ For\quad { t }^{ 8 }\quad 2q+3r=8\\ (i)\quad r=0,q=4\quad \therefore p=5\quad (from\quad equation\quad 1)\\ (ii)\quad r=2,q=1\quad \therefore p=6\\ \therefore \quad co-oefficient\quad of\quad { t }^{ 8 }\quad =\quad 9!\quad (\frac { { 1 }^{ 5 }{ 2 }^{ 4 }{ (-1) }^{ 0 } }{ (0)!(4)!(5)! } +\frac { { 1 }^{ 6 }{ 2 }^{ 1 }{ (-1) }^{ 2 } }{ (6)!(1)!(2)! } )\quad =\quad 2520.$

There are 2 ways to get a coefficient of $t^ 8$ , namely : $-t^3 \times -t^3 \times 2t^2$ and $2t^2 \times 2t^2 \times 2t^2 \times 2t^2$ , the coefficients being $2$ and $16$ respectively.

Considering the possible combinations of these, there are $\binom92 \binom71$ ways to arrive at $-t^3 \times -t^3 \times 2t^2$ and $\binom94$ ways to arrive at $2t^2 \times 2t^2 \times 2t^2 \times 2t^2$

Therefore coefficient of $t^8$ is : $\binom92 \binom71 \times 2 + \binom94 \times 16 = 2520$

It is basically a simple question of Multinomial Theorem .

@Sandeep Bhardwaj ,Is multinomial theorum ,riemann sum in JEE syllabus?