For any positive integer m,n with n≥m, let (nm)=nCm. Then (nm)+(n−1m)+(n−2m)+⋯+(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 Dn+1Cm+1 We know that, nCr= Coefficient of xr in (1+x)n ∴(nm)+(n−1m)+(n−2m)+⋯+(mm) =Coefficient ofxmin(1+x)n+Coefficient of xmin(1+x)n−1+⋯+Coefficient of xmin(1+x)m =Coefficient ofxmin[(1+x)n+(1+x)n−1+⋯+(1+x)m] =Coefficient ofxmin(1+x)m{(1+x)n−m+1−1(1+x)−1} =Coefficient ofxm+1in[(1+x)n+1−(1+x)m] =Coefficient ofxm+1in(1+x)n+1 =n+1Cm+1