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Question

In the expansion of (1+x)2(1+y)3(1+z)4(1+w)5, the sum of coefficients of the term of degree 12 is k. Then the value of k13 is

A
5
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B
6
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C
7
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D
8
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Solution

The correct option is C 7
Substituting y=x,z=x,w=x in the expression
Coefficient of x12 in (1+x)14 is
14C12=14!12!×2!=91
{ nCr=n!r!×(nr)!}
k13=9113=7

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