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Suppose a donuts shop has $20$ varieties of donuts. how many ways are there to choose at least two kinds of donuts in dozen donuts ?

please correct me:

this is combination with repetition question and we have total of $C\left(12+20-1,12\right)$ cases and we want to subtract it from the cases if we have one kind of donuts so this is $20$ so answer will be $C\left(31,12\right)-20$.

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Do you know why C(12+20-1,12) gives the number of combinations allowing repetitions? If you know why this formula works then you are not far from figuring out the number of ways of picking a single variety. – Montez Dec 19 '12 at 1:45
Yes, it’s fine. (But you’re not subtracting $\binom{31}{12}$ from $20$; you’re subtracting $20$ from $\binom{31}{12}$. Subtract $a$ from $b$ means $b-a$.) – Brian M. Scott Dec 19 '12 at 1:58
up vote 2 down vote accepted

That sounds correct. It's certainly easier to subtract the outcomes you don't want (selecting only one type) than to add up the ones you do want (selecting 2, 3, ..., 12 types).

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