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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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    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 $b-a$.)2012-12-19

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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).