$\ X_i $ is discrete random variable, Compute $\ \sum X_i = 97 $ Let $\ X_1, X_2, , \dots , X_{10} $ be a discrete random variable with uniform distribution between $\ 0 $ to $\ 10 $. Compute $\ P\{ \sum_{i=1}^{10} \ X_i = 97 \} $, the variables are independent.
My attempt:
So I can either get 97 by having nine 10's and a 7, eight 10's and 9 & 8 or seven 10's and three 9's ?
$$\ P\{ \sum_{i=1}^{10} X_i = 97 \} = {10 \choose 1 } \cdot \frac{1}{11}^{10} + {10 \choose 2} \cdot \frac{1}{11}^{10} + {10 \choose 3} \cdot \frac{1}{11}^{10} $$

Your solution is almost correct.
The correct solution is:$$11^{-10}\left(\frac{10!}{9!1!}+\frac{10!}{8!1!1!}+\frac{10!}{7!3!}\right)$$
Observe that $8$ and $9$ are distinct numbers. 
So e.g. $(10,10,9,10,10,10,10,10,8,10)$ and $(10,10,8,10,10,10,10,10,9,10)$ are distinct possibilities.
That is why $\binom{10}{2}=\frac{10!}{8!2!}$ must be replaced by $\frac{10!}{8!1!1!}$.
