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Question
For any positive integer m,n with n≥m, let (nm)=nCm. Then
(nm)+(n−1m)+(n−2m)+⋯+(mm) is equal to
For any positive integer m,n with n≥m, let (nm)=nCm. Then
(nm)+(n−1m)+(n−2m)+⋯+(mm) is equal to
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Solution
The correct option is C n+1Cm+1
We know that,
nCr= Coefficient of xr in (1+x)n
∴(nm)+(n−1m)+(n−2m)+⋯+(mm)
=Coefficient of xm in (1+x)n+ Coefficient of xm in (1+x)n−1 +⋯+ Coefficient of xm in (1+x)m
=Coefficient of xm in [(1+x)n+(1+x)n−1+⋯+(1+x)m]
=Coefficient of xm in (1+x)m{(1+x)n−m+1−1(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
We know that,
nCr= Coefficient of xr in (1+x)n
∴(nm)+(n−1m)+(n−2m)+⋯+(mm)
=Coefficient of xm in (1+x)n+ Coefficient of xm in (1+x)n−1 +⋯+ Coefficient of xm in (1+x)m
=Coefficient of xm in [(1+x)n+(1+x)n−1+⋯+(1+x)m]
=Coefficient of xm in (1+x)m{(1+x)n−m+1−1(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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