The answer depends on if you can choose the same kind of donuts
more then once. Or in other words, is repetition permitted.
If you can only choose the same kind of donuts only once, it is
a 21 choose 12 problem:
C(n,k) = n! / (k! (n - k)!)
C(21, 12) = 21! / (12! (21 - 12)!)
= 21! / (12! (9)!)
= 293,930
If you can choose the same kind of donuts more then once, it is
a combination with repetition problem.
P(n+k-1,k) = (n+k-1)! / (k! (n-1)!)
or put it into C(n,k) with n+k-1 as 21 + 12 - 1 = 32 and k as 12
so
C(21+12-1,12)
= C(32, 12)
= 32! / (12! (32 - 12)!)
= 32! / (12! (20)!)
= 225,792,840