The correct option is B 52C4
=47C4+52−1C3+52−2C3+52−3C3+52−4C3+52−5C3
=47C4+51C3+50C3+49C3+4C3+47C3
=(47C4+47C3)+4C3+49C3+50C3+51C3
=(48C4+48C3+49C3+50C3+51C3
(∵nCr+nCr−1=n+1Cr)
=(49C4+49C3)+50C3+51C3)(∵nCr+nCr−1=n+1Cr)
=(50C4+50C3)+51C3)(∵nCr+nCr−1=n+1Cr)
=(51C4+51C3(∵nCr+nCr−1=n+1Cr)
=52C4(∵nCr+nCr−1=n+1Cr)