Indian Institute of Foreign Trade 2010 Solved Paper

We can write $\Rightarrow {\left(1-{\mathit{x}}^6\right)}^4=\left(1-4{\mathit{x}}^6+6{\mathit{x}}^{12}-4{\mathit{x}}^{18}+{\mathit{x}}^{24}\right)$
Therefore, we can say
We have to find out coefficient of
We can use binomial expansion for negative coefficients. Therefore, coefficient of ${\mathit{x}}^{12}$ in ${\left(1-\mathit{x}\right)}^{-4}$
$\Rightarrow \frac{\left(-4\right)\times \left(-4-1\right)\left(-4-2\right)\times ...\times \left(-4-11\right)}{12!}$
$\Rightarrow \frac{15!}{12!\times 3!}$
$\Rightarrow \frac{15\times 14\times 13}{3\times 2\times 1}$
$\Rightarrow 455$
Similarly, coefficient of ${\mathit{x}}^6$ in ${\left(1-\mathit{x}\right)}^{-4}$
$\Rightarrow \frac{9!}{6!\times 3!}$
$\Rightarrow \frac{7\times 8\times 9}{3\times 2\times 1}$
$\Rightarrow 84$
Coefficient of ${\mathit{x}}^0$ in ${\left(1-\mathit{x}\right)}^{-4}$ is 1
Therefore, we can say that the coefficient of ${\mathit{x}}^{12}$ in the expansion of Hence, option C is the correct answer.
Alternative Solution :${\left(1-{\mathit{x}}^6\right)}^4={\left(1-\mathit{x}\right)}^4{\left(1+\mathit{x}+{\mathit{x}}^2+{\mathit{x}}^3+{\mathit{x}}^4+{\mathit{x}}^5\right)}^4$
Hence we need to find co-eff of ${\mathit{x}}^{12}$ in ${\left(1+\mathit{x}+{\mathit{x}}^2+{\mathit{x}}^3+{\mathit{x}}^4+{\mathit{x}}^5\right)}^4$
This will be equal to number of integral solutions for
a is the power of x from the first expression, b is the power of x
Lets find the set of values for (a,b,c,d)
$\left(5,5,2,0\right)$ ⇒ Number of ways of arranging $=\frac{4!}{2!}=12$
$\left(5,5,1,1\right)$⇒ Number of ways of arranging $=\frac{4!}{\left(2!\times 2!\right)}=6$
$\left(5,4,3,0\right)$ ⇒ Number of ways of arranging$=4!=24$
$\left(5,4,2,1\right)$ ⇒ Number of ways of arranging $=4!=24$
$\left(5,3,2,2\right)$⇒ Number of ways of arranging $=\frac{4!}{2!}=12$
$\left(5,3,3,1\right)$⇒ Number of ways of arranging $=\frac{4!}{2!}=12$
$\left(4,4,4,0\right)$⇒ Number of ways of arranging$=\frac{4!}{3!}=4$
$\left(4,4,3,1\right)$ ⇒ Number of ways of arranging $=\frac{4!}{2!}=12$
$\left(4,4,2,2\right)$⇒ Number of ways of arranging $=\frac{4!}{\left(2!\times 2!\right)}=6$
$\left(4,3,3,2\right)$ ⇒ Number of ways of arranging $=\frac{4!}{2!}=12$
$\left(3,3,3,3\right)$ ⇒ Number of ways of arranging $=\frac{4!}{4!}=1$
Hence the co-eff of ${\mathit{x}}^{12}=24\times 2+6\times 2+4+1=125$