I cant figure out this bonus question for stats hw

How many terms are there for

(x+y+z)^80

This is the number of triples (a,b,c) such that a+b+c = 80, with a b and c at least 0. When you have a pair (a,b) of sum at most 80, you have only (exactly) 1 choice for c, so that's the number of pairs (a,b) of sum at most 80. That is \sum_{a=0 to 80}(81-a). Which is 81*81 - 81*80/2. So 81*41. errr 3321 ?

This is equivalent to the number of nonnegative integer valued vectors (x1,x2,x3,...xr) satisfying x1 +x2+x3+...+xr = n

which is just (n+r -1)C(r-1). I'll prove it shortly.


So we just plug in r=3 and n=80 to give us 3321 which agrees with Jean's result.

Here is a combinatorial proof of the above.

Take r-1 O's and n |'s. There are clearly (n+r-1)C(r-1) different combinations of these letters. Now the l's divide the O's and tells us the values of the xi's.


i.e., lll0000 implies that x1=3, x2=0, x3=0, x4=0,x5=0
|0||000 implies that x1=1,x2=2,x3=x4=x5=0

Yeah, thanks for the general formula. It's much better than my messy summation You're making me feel a bit ashamed of myself, as the combinatorial bijection is very simple (and cute) ...