Answer is C(22,2). But I want to learn the logic of it.
If we select $N$ items from $K$ different choices (where repetition is allowed), the number of ways that can be done is $\dbinom{N+K-1}{N}$.
In expanding $( x+ y+ z)^{20}$ each term looks like $Ax^jy^kz^m$ where $j+k+m=20$.
So we are choosing $20$ ones from $3$ kinds: $\dbinom{22}{2}$.
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