Let X = 10 C 1 2 + 2 10 C 2 2 + 3 10 C 3 2 + … + 10 10 C 10 2 , where 10 C r ,   r ∈ 1,2 ,…

Let X=10C12+210C22+310C32++1010C102 , where 10Cr, r1,2,.,10 denote binomial coefficients. Then, the value of 11430X is ______ .

Solution

X=r=0nr.nCr2;n=10
X=n.r=0nnCr.  n-1Cr-1
X=n.r=1nnCn-r.  n-1Cr-1
X=n. 2n-1Cn-1;n=10
X=10. 19C9
X1430=1143.19C9
=646

Asked in: JEE Advanced 2018 (Paper 2)

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