What is the coefficient of x^9 y^16 in the expansion of (7x+21y)^25 binomial(n, k) x^(n-k) y^k 
Given newtons binomial theorem.
I believe the answer is 
n = 25
k = 16
binomial(25, 16) 7^(9) 21^16 

Yes true. By binomial theorem we have
$$(7x+21y)^{25}=\sum_{k=0}^{25}{25\choose k}(7x)^k(21y)^{25-k}$$
so for $k=9$ the term is
$${25\choose 9}7^9\times (21)^{16} x^9 y^{16}$$
