Iift 2010 Question 25
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IIFT 2010
(1 − x6)4 4C0(1)4(x6)0 4C1(1)3(x6)1 4C2(1)2(x6)2 4C3(1)1(x6)3 4C4(1)0(x6)4
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