Finding the Binomial Coffecient

Question

Is there a way to find the binomial coefficient of $x^{14}$ in $$(x^0+x^1+x^2+x^3+x^4)^6$$

I tried to use sum of G.P form but it did'nt help.

Answer 1

HINT:

$$\sum_{r=0}^4 x^r=\frac{1-x^5}{1-x}$$

$$\implies(x^0+x^1+x^2+x^3+x^4)^6=(1-x^5)^6(1-x)^{-6}$$

Now $(1-x^5)^6=1-\binom61x^5+\binom62(x^5)^2-\binom63(x^5)^3+\cdots+x^{30}$

and $(1-x)^{-6}=1+(-6)(x)+\frac{(-6)(-6-1)}{2!}x^2+\frac{(-6)(-6-1)(-6-2)}{3!}x^3+\cdots$