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I think this is answer 1 5 5 10 20 10 10 30 30 10 5 20 30 20 5 1 5 10 10 5 1

Another way to expand (a+b+c)n is using Pascal's Triangle.

It doesn't matter for i+j+k=n is over or under the summation symbol. It means how many ways to make i+j+k=n, where i,j,k≥ 0. For example: (a+b+c)3 i+j+k=3 How many ways to make i+j+k=3? 3+0+0=3 => i=3,j=0,k=0 0+3+0=3 => i=0,j=3,k=0 0+0+3=3 => i=0,j=0,k=3 1+1+1=3 => i=1,j=1,k=1...

When (w+x+y+z)23 is expanded and terms collected, how many different terms will there be? from pascal's triangle you can find \frac{26!}{23!3!}= 5200 diffenent terms.