Indian Institute of Foreign Trade 2010 Solved Paper

We can write $\Rightarrow (1-x^{6})^{4}=(1-4x^{6}+6x^{12}-4x^{18}+x^{24})$Therefore, we can sayWe have to find out coefficient of $x^{12}, x^{6}, x^{0}\;\text{in}\;(1-x)^{-4}$We can use binomial expansion for negative coefficients. Therefore, coefficient of $x^{12}$ in $(1-x)^{-4}$$\Rightarrow \frac{(-4)\times(-4-1)(-4-2)\times\ldots\times(-4-11)}{12!}$$\Rightarrow \frac{15!}{12!\times 3!}$$\Rightarrow \frac{15\times14\times13}{3\times2\times1}$$\Rightarrow 455$Similarly, coefficient of $x^{6}$ in $(1-x)^{-4}$$\Rightarrow \frac{9!}{6!\times 3!}$$\Rightarrow \frac{7\times8\times9}{3\times2\times1}$$\Rightarrow 84$Coefficient of $x^{0}$ in $(1-x)^{-4}$ is 1Therefore, we can say that the coefficient of $x^{12}$ in the expansion of Hence, option C is the correct answer.Alternative Solution :$(1-x^{6})^{4}=(1-x)^{4}(1+x+x^{2}+x^{3}+x^{4}+x^{5})^{4}$Hence we need to find co-eff of $x^{12}$ in $(1+x+x^{2}+x^{3}+x^{4}+x^{5})^{4}$This will be equal to number of integral solutions for $a+b+c+d=12, 0 \le a,b,c,d \le 4$a is the power of x from the first expression, b is the power of xLets find the set of values for (a,b,c,d)$(5,5,2,0)$ ⇒ Number of ways of arranging $=\frac{4!}{2!}=12$$(5,5,1,1)$⇒ Number of ways of arranging $=\frac{4!}{2!\times 2!}=6$$(5,4,3,0)$ ⇒ Number of ways of arranging$=4!=24$$(5,4,2,1)$ ⇒ Number of ways of arranging $=4!=24$$(5,3,2,2)$⇒ Number of ways of arranging $=\frac{4!}{2!}=12$$(5,3,3,1)$⇒ Number of ways of arranging $=\frac{4!}{2!}=12$$(4,4,4,0)$⇒ Number of ways of arranging$=\frac{4!}{3!}=4$$(4,4,3,1)$ ⇒ Number of ways of arranging $=\frac{4!}{2!}=12$$(4,4,2,2)$⇒ Number of ways of arranging $=\frac{4!}{2!\times 2!}=6$$(4,3,3,2)$ ⇒ Number of ways of arranging $=\frac{4!}{2!}=12$$(3,3,3,3)$ ⇒ Number of ways of arranging $=\frac{4!}{4!}=1$Hence the co-eff of $x^{12}=24\times2+6\times2+4+1=125$