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B
m+nCm+r
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C
m+nCr
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D
0
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Solution
The correct option is Cm+nCr Given equation ismCr+mCr−1nC1...mC1nCr−1+nCr. Vandermonde's identity states that. (m+nk)=∑rk=0(mk)(nr−k) Hence by application of Vandermonde's identity. mCr+mCr−1nC1...mC1nCr−1+nCr =m+nCr