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Question

The value of 37C0+47C1+57C2+107C7 is ------------

A
10(2)6
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B
13(2)7
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C
14(2)6
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D
13(2)6
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Solution

The correct option is A 10(2)6
Given that,
37C0+47C1+57C2+...............107C7
tn=(r+3)nCr
We know that,
[rnCr=n.n1Cr]
Sn=7r=0+n=7r=0rnCr+37r=0nCr
=7r=076Cr+37r=07Cr
=7×26+3×27
=13(2)6
Then,
Option A is correct answer.

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