The coefficient of $x^{12}$ in $(x^{3}+x^{4}+x^{5}+x^{6}+…)^{3} $ is ___________.
$\Rightarrow$ $(x^{3}+x^{4}+x^{5}+x^{6}+…...)^{3}$
$\Rightarrow$ $[x^{3}(1+x+x^{2}+x^{3}+x^{4}+…...)]^{3}$
$\Rightarrow$ $[x^{3}(1+x+x^{2}+x^{3}+x^{4}+…...)]^{3}$
$\Rightarrow$ $x^{9}[1+ x+x^{2}+x^{3}+x^{4}+…]^{3}$ ------------->(1)
Let generating sequence $(1,1,1,1,1,1,1......................)$
write the generating function:
$G(x) = 1+x+x^{2} + x^{3} + x^{4} +............$
$G(x) = \frac{1}{(1-x)}$ [Sum of infinite series]
Put this value in equation(1)
$\Rightarrow$ $x^{9}[\frac{1}{(1-x)}]^{3}$
$\Rightarrow$ $x^{9}.\frac{1}{(1-x)^{3}}$
$\Rightarrow$ $x^{9}.(1-x)^{-3}$
Apply Binomial theorem
$(a+b)^{n}=\sum_{k=0}^{n}\binom{n}{k}(a)^{n-k}.(b)^{k}$
$\Rightarrow$$(a+b)^{-n}=\sum_{k=0}^{\infty}\binom{-n}{k}(a)^{-n-k}.(b)^{k}$
$\Rightarrow$$(a+b)^{-n}=\sum_{k=0}^{\infty}\binom{n+k-1}{k}(-1)^{k}(a)^{-n-k}.(b)^{k}$
Now i find the $(1-x)^{-3}=\sum_{k=0}^{\infty}\binom{3+k-1}{k}(-1)^{k}(1)^{-3-k}.(-x)^{k}$
$\Rightarrow$$x^{9}.(1-x)^{-3}= x^{9}.\sum_{k=0}^{\infty}\binom{3+k-1}{k}(-1)^{k}.(-x)^{k}$
Put k =3,we get $x^{12},$
$\Rightarrow$ $x^{9}.\binom{3+3-1}{3}(-1)^{3}.(-x)^{3}$
$\Rightarrow$ $x^{9}.\binom{5}{3}(-1)^{3}.(-1)^{3}.(x)^{3}$
$\Rightarrow$ $x^{9}.\binom{5}{3}(-1).(-1).(x)^{3}$
$\Rightarrow$ $x^{9}.\binom{5}{3}.x^{3}$
$\Rightarrow$ $\binom{5}{3}.x^{3}x^{9}.$
Coefficient of $x^{12}\: \text{is} : \binom{5}{3}$
$\Rightarrow \frac{5!}{(5-3)!.3!}$
$\Rightarrow \frac{5!}{2!.3!}$
$\Rightarrow \frac{5.4.3!}{2!.3!}$
$\Rightarrow \frac{5.4}{2.1 }$
$\Rightarrow10$
So, the correct answer is 10.