7. binomial theoram — ગણિત ધોરણ 11 સાયન્સ — Question

Applying to the above question we get $T_{r+1}=(-1)^{r+10} C_{r} x^{10-r} 2^{r-10} 3^{r} x^{-2 r}$

$=(-1)^{r \cdot 10} C_{r} x^{10-3 r} 2^{r-10} 3^{r} \ldots$

For the coefficient of $x^{4}$

$10-3 r=4$

$6=3 r$

$r=2$

Substituting in (i) we get $T_{3}={ }^{10} C_{2} x^{4} 2^{-8} 3^{2}$

$=\frac{10.9 .3^{2}}{2 ! 2^{8}}$

$=\frac{405}{256}$