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Question

For any positive integer m,n with nm, let (nm)=nCm. Then
(nm)+(n1m)+(n2m)++(mm) is equal to

A
nCm
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B
n+1Cm
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C
nCm+1
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D
n+1Cm+1
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Solution

The correct option is D n+1Cm+1
We know that,
nCr= Coefficient of xr in (1+x)n
(nm)+(n1m)+(n2m)++(mm)
=Coefficient of xm in (1+x)n+ Coefficient of xm in (1+x)n1 ++ Coefficient of xm in (1+x)m
=Coefficient of xm in [(1+x)n+(1+x)n1++(1+x)m]
=Coefficient of xm in (1+x)m{(1+x)nm+11(1+x)1}
=Coefficient of xm+1 in [(1+x)n+1(1+x)m]
=Coefficient of xm+1 in (1+x)n+1
=n+1Cm+1

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