Find the value of C0C3+C1C4+.....Cn−3Cn, when Cr=nCr
We know nC3=nCn−3,nC4=nCn−4.....
⇒sum=nC0nCn−3+nC1nCn−4....nCn−3nC0
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 xn−3.
(1+x)n(1+x)n=(nC0+nC1x+nC2x2.....nCnxn)(nC0+nC1x+nC2x2.....nCnxn)
Coefficient of xn−3=nC0×nCn−3+nC1nCn−4.....nCn−3nC0
This is same as the required. But we can calculate the coefficient of xn−3 easily.
Sum = coefficient of xn−3 in (1+x)n(1+x)n
= coefficient of xn−3 in (1+x)2n
=2nCn−3