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Question

nm1(mk1(mpk nCm.mCp.pCk))=

A
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B
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Solution

The correct option is B
nm=1nCm(mk=1(mp=km!p!(mp)!.p!k!(pk)!))
=nm=1nCm(mk=1(mp=kmkCpk)m!k!(mk)!)
=nm1nCm(mk12mk.mCk)
=nm1nCm((1+2)m2m)=nm1(nCm3mnCm2m)
=(1+3)n1(1+2)n+1=4n3n

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