Find the coefficient of x5 in (x + 3)8

It is known that $(r+1)^{\text {th }}$ term,$\left(T_{r+1}\right)$, in the binomial expansion of $(a+b)^{n}$ is given by $T_{r+1}={ }^{n} C_{r} a^{n-r} b^{r}$.

Assuming that $x^{5}$ occurs in the $(r+1)^{\text {th }}$ term of the expansion $(x+3)^{8}$, we obtain

$T_{r+1}={ }^{8} C_{r}(x)^{8-r}(3)^{r}$

Comparing the indices of $x$ in $x^{5}$ and in $T_{r+1}$, we obtain

$r=3$

Thus, the coefficient of $x^{5}$ is $^{8} \mathrm{C}_{3}(3)^{3}=\frac{8 !}{3 ! 5 !} \times 3^{3}=\frac{8 \cdot 7 \cdot 6 \cdot 5 !}{3 \cdot 2.5 !} \cdot 3^{3}=1512$