Counting donut problems By using the permutation and combination techniques, i have attempt to solve this problem and i would like to know if where i did it wrong
how many ways to choose $12$ donuts from a store that offers thirty varieties?
here is my computation
$\Large{\binom{12+30-1}{12}= \binom{41}{12}= \frac{41!}{12!(41-12)!}}$ by using the formula $\Large\binom{n}{k}=\frac{n!}{k!(n-k)!}$
Could you tell me if i did it wrong or right??
Thank you
sincerely

Your answer is correct - assuming the following things that you did not state:
1) The order in which the donuts are chosen does not matter.
2) You are allowed to choose several donuts of the same type.
If order didn't matter and we didn't allow repetition, then the correct answer would be $\binom{30}{12}$.
If we allowed repetition and order mattered, then the correct answer would be $30^{12}$.
And if the order mattered and repetitions were not allowed, then the correct answer would be $(30)_{12}$; if you haven't seen this notation before, it is the falling factorial, and is defined by
$$
(a)_b=a(a-1)(a-2)\cdots(a-(b-1))=\frac{a!}{(a-b)!}.
$$
