Let F(n,m) = the number of terms in a polynomial in n variables of order m.
Then F(1, 2) = 3 since you have
A + Bx + Cx^2
(where x^2 means "x squared".)
and F(1,3) = 4:
A + Bx + Cx^2 + Dx^3.
Also, F(2, 1) = 3:
A + Bx + Cy
and F(3,1) = 4:
A + Bx + Cy + Dz.
SomeElementaryReasoning will convince you that:
F(n, m) = F(n-1, m) + F(n-1, m-1) + F(n-1, m-2) + ... + F(n-1, 2) + F(n-1, 1) + 1
and SomeBasicThoughts will give you:
F(n,0) = 1
F(n,1) = n+1
F(1,m) = m+1
and let's define:
F(0,m) = 1
(Peter Balaam, 29/07/2003)
F(n,m) = F(n-1,m) + F(n, m-1)
(Peter Balaam, 29/07/2003)
F(n,m) = F(m,n)
(Marieke Rohde, 23/11/2004)
F(n,m) = (n+m)! / (m!n!)
And that about wraps that up.