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Question

The value of 47C4+5r=152rC3=

A
52C2
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B
52C3
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C
52C4
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D
52C5
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Solution

The correct option is B 52C4
We know that,
nCr+nCr1

=n!r!(nr)!+n!(r1)!(nr+1)!

=n!(nr+1)r!(nr)!(nr+1)+n!r(r1)!(nr+1)!r

=n!(nr+1)r!(nr+1)!+n!r(nr+1)!r!

=n!r!(nr+1)!(nr+1+r)

=n!r!(nr+1)!(n+1)

=(n+1)!r!(nr+1)!

=n+1Cr

nCr+nCr1=n+1Cr
The equation is: (47C4+47C3)+48C3+49C3+50C3+51C3
=(48C4+48C3)+49C3+50C3+51C3
=(49C4+49C3)+50C3+51C3
=(50C4+50C3)+51C3
=(51C4+51C3)
=52C4
Hence, option 'C' is correct.

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