How many ways, can sum be equal to 12 of 3 dice? I've tried solving it in the below method,
$$X_1+X_2+X_3 =12$$
Using Stars and Bars method, we've to restrict one star for all the three entries,
$$X_1+X_2+X_3=9$$
Now we have $\binom{11}{9}$ possibilities where each dice value will be atleast or greater than $1$. 
Now I've to remove the entries which have $7$ or more than $7$ entries ($X_i\ge 7$) from the $\binom{11}{9}$ possibilities. So equation can be further deduce to
$$X_1+X_2+X_3=2\qquad  (=9-7)$$
So there are $\binom{4}{2}$ possibilities where the $X_i$ values will not be $7$ or more than $7$.
$$\text{Final ans :}\: \binom{11}{9} - \binom{4}{2}$$
But the answer is given as $\binom{11}{9}-\binom{5}{3}$.
Can someone please explain where my reasoning was wrong?

There are two small errors I can see:


*

*Where you subtract the ways in which each $X_i$ can be greater than or equal to $7$ you subtract $7$ when you should subtract $6$ giving $X_1+X_2+X_3=3$, this is because you have already put $1$ star into each bin at the start.

*Also you have only subtracted $1$ such case when in fact there are $\binom{3}{1}=3$ cases, i.e. The number of ways you can choose $1$ of the $X_i$ to be $\ge 7$ out of the $3$ $X_i$s. 


So your final answer should be
$$\binom{11}{9}-\binom{3}{1}\binom{5}{2}=25\tag{Answer}$$
The answer you have been given is incorrect and I don't know how it was arrived at (perhaps they forgot the $3$ cases too).
