Solutions in non negative and positive intergers

Hi, I've been going over some problems that I don't have answers too and just need some verification that what im doing is right!!

Find the number of solutions in a) non negative and b) positive intergers for $\displaystyle a+b+c+d=12$

For a) I have $\displaystyle \binom{12+4-1}{4-1} = \binom{15}{3} = 455$

and b) $\displaystyle \binom{8+4-1}{4-1} = \binom{11}{3} = 165$

I think/hope that is right but I've got no way of checking!! Cheers for any help!