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B
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C
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D
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Solution
The correct option is B ∑nm=1nCm(∑mk=1(∑mp=km!p!(m−p)!.p!k!(p−k)!)) =∑nm=1nCm(∑mk=1(∑mp=km−kCp−k)m!k!(m−k)!) =∑nm−1nCm(∑mk−12m−k.mCk) =∑nm−1nCm((1+2)m−2m)=∑nm−1(nCm3m−nCm2m) =(1+3)n−1−(1+2)n+1=4n−3n