Find the coefficient of $$x^{12}$$ in the expansion of $$(1 - x^{6})^{4}(1 - x)^{-4}$$

We can write $$(1 - x^{6})^{4}$$ = $$4C0(1)^4(x^6)^0$$ - $$4C1(1)^3(x^6)^1$$ + $$4C2(1)^2(x^6)^2$$ - $$4C3(1)^1(x^6)^3$$+ $$4C4(1)^0(x^6)^4$$

$$\Rightarrow$$ $$(1 - x^{6})^{4}= (1-4x^6+6x^{12}-4x^{18}+x^{24})$$

Therefore, we can say

$$(1 - x^{6})^{4}(1 - x)^{-4}=(1-4x^6+6x^{12}-4x^{18}+x^{24})*(1 - x)^{-4}$$

We have to find out coefficient of $$x^{12}$$, $$x^6$$, $$x^0$$ in $$(1 - x)^{-4}$$.

We can use binomial expansion for negative coefficients. Therefore, coefficient of $$x^{12}$$ in $$(1 - x)^{-4}$$

$$\Rightarrow$$ $$\dfrac{(-4)*(-4-1)*(-4-2)* ... *(-4-11)}{12!}$$

$$\Rightarrow$$ $$\dfrac{15!}{12!*3!}$$

$$\Rightarrow$$ $$\dfrac{15*14*13}{3*2*1}$$

$$\Rightarrow$$ $$455$$

Similarly, coefficient of $$x^6$$ in $$(1 - x)^{-4}$$

$$\Rightarrow$$ $$\dfrac{(-4)*(-4-1)*(-4-2)* ... *(-4-5)}{6!}$$

$$\Rightarrow$$ $$\dfrac{9!}{6!*3!}$$

$$\Rightarrow$$ $$\dfrac{7*8*9}{3*2*1}$$

$$\Rightarrow$$ $$84$$

Coefficient of $$x^0$$ in $$(1 - x)^{-4}$$ is 1. 

Therefore, we can say that the coefficient of $$x^{12}$$ in the expansion of $$(1 - x^{6})^{4}(1 - x)^{-4}$$ = 455+(-4*84)+(1*6) = 125. Hence, option C is the correct answer. 

Alternative Solution:

$$(1-x^6)^4 = (1-x)^4(1+x+x^2+x^3+x^4+x^5)^4$$

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

Hence we need to find coeff of  $$x^{12}$$ in $$(1+x+x^2+x^3+x^4+x^5)^4$$ =  $$(1+x+x^2+x^3+x^4+x^5)\times$$$$(1+x+x^2+x^3+x^4+x^5)\times$$$$(1+x+x^2+x^3+x^4+x^5)\times$$$$(1+x+x^2+x^3+x^4+x^5)$$

This will be equal to number of integral solutions for a + b + c + d = 12, 0<=a,b,c,d<=4

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)

(5,5,2,0) => Number of ways of arranging = 4!/2! = 12

(5,5,1,1) => Number of ways of arranging = 4!/(2!*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 = 4!/2! = 12

(5,3,3,1) => Number of ways of arranging = 4!/2! = 12

(4,4,4,0) => Number of ways of arranging = 4!/3! = 4

(4,4,3,1) => Number of ways of arranging = 4!/2! = 12

(4,4,2,2) => Number of ways of arranging = 4!/(2!*2!) = 6

(4,3,3,2) => Number of ways of arranging = 4!/2! = 12

(3,3,3,3) => Number of ways of arranging = 4!/4! = 1

Hence the coeff of $$x^{12}$$ = 24*2 + 12*5 + 6*2 + 4 + 1 = 125