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]
(646)
∑r=0nr(Crn)2=n∑r=0nCrnCr-1n-1
=n∑r=1nCn-rnCr-1n-1=nCn-12n-1
Now, X=(C110)2+2(C210)2+3(C310)2+⋯+10(C1010)2
=∑n=010r(Cr10)2=10 C919
∴ X1430=1143 C919=646