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Question

Find the value of C0C3+C1C4+.....Cn3Cn, when Cr=nCr


A

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B

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C

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D

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Solution

The correct option is A


We know nC3=nCn3,nC4=nCn4.....
sum=nC0nCn3+nC1nCn4....nCn3nC0
The sum of the subscripts is same in each term = n - 3
Consider the expansion of (1+x)n(1+x)n and find the coefficient of xn3.
(1+x)n(1+x)n=(nC0+nC1x+nC2x2.....nCnxn)(nC0+nC1x+nC2x2.....nCnxn)
Coefficient of xn3=nC0×nCn3+nC1nCn4.....nCn3nC0
This is same as the required. But we can calculate the coefficient of xn3 easily.
Sum = coefficient of xn3 in (1+x)n(1+x)n
= coefficient of xn3 in (1+x)2n
=2nCn3


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