We can write
(1−x6)4=4Co0(1)4(x)6−4C1(1)3(¯y6)1+4C2(1)2(x6)2−4C3(1)1(x6)3+4C4(1)0(x6)4
⇒(1−x6)4=(1−4x6+6x12−4x18+x24)Therefore, we can say
(1−x6)4(1−x)−4=(1−4x6+6x12−4x18+x24)×(1−x)−4
We have to find out coefficient of
x12,x6,x0‌in‌(1−x)−4We can use binomial expansion for negative coefficients. Therefore, coefficient of
x12 in
(1−x)−4⇒| (−4)×(−4−1)(−4−2)×...×(−4−11) |
| 12! |
⇒⇒⇒455Similarly, coefficient of
x6 in
(1−x)−4⇒⇒⇒84Coefficient of
x0 in
(1−x)−4 is 1
Therefore, we can say that the coefficient of
x12 in the expansion of
(1−x6)4(1−x4)=455+(−4×84)+(1×6)=125
Hence, option C is the correct answer.
Alternative Solution :(1−x6)4=(1−x)4(1+x+x2+x3+x4+x5)4⇒(1−x6)4(1−x)−4=(1+x+x2+x3+x4+x5)
Hence we need to find co-eff of
x12 in
(1+x+x2+x3+x4+x5)4=(1+x+x+x2+x3+x4+x5)(1+x+x2+x3+x4+x5)(1+x+x2+x3+x4+x5)
This will be equal to number of integral solutions for
a+b+c+d=12,0<=a,b,c,d<=4a 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
==12(5,5,1,1)⇒ Number of ways of arranging
==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
==12(5,3,3,1)⇒ Number of ways of arranging
==12(4,4,4,0)⇒ Number of ways of arranging
==4(4,4,3,1) ⇒ Number of ways of arranging
==12(4,4,2,2)⇒ Number of ways of arranging
==6(4,3,3,2) ⇒ Number of ways of arranging
==12(3,3,3,3) ⇒ Number of ways of arranging
==1Hence the co-eff of
x12=24×2+6×2+4+1=125