Question

Coefficient of $\left(x^3{y}^4{z}^2\right)$ in $(2x-3y+4x{)}^9$

Answer 1

$(2x-3y+4z{)}^9=[(2x-3y)+4z{]}^9$

General term, $T_{rH}={}^9{C}_r(2x-3y{)}^r(4z{)}^{9-r}$

General term for $(2x-3y{)}^r={}^r{C}_s(2x{)}^s(-3y{)}^{r-s}$

$\Rightarrow T_{rH}={}^s{C}_r\sum _{s=0}^r{}^r{C}_s\left(2x{)}^s\right(-3y{)}^{r-s}(4z{)}^{9-r}$

For $x^3{y}^4{z}^2,9-r=2\Rightarrow r=7,s=3$

$\Rightarrow$ Coefficient $={}^9{C}_7\cdot {}^7{C}_3\left(2{)}^3\right(-3{)}^4(4{)}^2$

$=\frac{9.8}{2}\times \frac{\mathrm{7.6.5}}{6}.8.81.16=36\times 35\times 8\times 81\times 16$

$=13063680$.