(In reply to
re: Trivial solution? by Ady TZIDON)
So if we have (a + b + c)^n, then there will be all of the following terms in the expansion:
a^n
b^n
c^n
a^(n1)*b
a^(n1)*c
a^(n2)*b*c
a^(n2)*b^2
etc.
Basically, every combination of a, b, and c such that their exponents add up to n. There are (n+2) choose 2 such combinations, since we have to order the n powers and 2 partitions between the three components of the polynomial.
So if a polynomial of m terms is raised to the nth power, the number of terms in the expansion will be (n+m1) choose (m1).
So we have the number of terms in the expansion of P1, N(P1) = (n+m1) choose (n1) = (n+m1)! / m!(n1)!
and N(P2) = (n+m1) choose (m1) = (n+m1)! / n!(m1)!
This simplifies nicely to N(P1) = N(P2) * n / m
So if 2 < n < m, then n/m < 1 and thus N(P1) < N(P2) for all n,m (which agrees with my "trivial" observation).
Edited on July 16, 2014, 3:46 pm

Posted by tomarken
on 20140716 14:05:56 