Q.

Let X=(C110)2+2(C210)2+3(C310)2++10(C1010)2, where Cr10, r{1,2,,10} denote binomial coefficients. Then, the value of 11430X is ______ .      [2018]


Ans.

(646)

r=0nr(Crn)2=nr=0nCrnCr-1n-1

=nr=1nCn-rnCr-1n-1=nCn-12n-1

Now, X=(C110)2+2(C210)2+3(C310)2++10(C1010)2

=n=010r(Cr10)2=10C919

 X1430=1143C919=646