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Question

Find the value of 47C4+5r=1 52rC3=?


Solution

The correct option is B
47C4+5r=152rC3=47C4+51C3+50C3+49C3+48C3+47C3
We know,
nCr+nCr1=n+1Cr
Simplify it by taking two terms at a time.
47C4+5r=152rC3=47C4+47C3+48C3+49C3+50C3+51C3
=48C4+48C3+49C3+50C3+51C3
=49C4+49C3+50C3+51C3
=50C4+50C3+51C3
=51C4+51C3
=52C4

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