We have to evaluate this: $$\sum_{r=0}^{10} r \binom{10}{r} 3^r (-2)^{10-r}$$
What i started doing is expand the binomial coefficient as:
$$r\binom{10}{r} = \dfrac{10!}{(10-r)!\ (r-1)!} = 10 \binom{9}{r-1}$$
Also we take $3$ common to get the sum:
$$10\cdot3\sum_{r=1}^{10} \binom{9}{r-1} 3^{r-1} (-2)^{9-(r-1)}$$
which is same as
$$30(3-2)^9 = 30$$
But answer is given as $\boxed{15}$?