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 ______ .


Ans.

(646)

X=∑r=110r(Cr10)2=∑r=110r·Cr10·Cr10

=10∑r=110Cr-19·C10-r10=10·C919

∴  X1430=10·C9191430=C919143=646