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Binomial theorem for positive integral Index

Question

Let X= 10C12+2 10C22+3 10C32++10 10C102 where 10Cr,r{1,2,,10} denote binomial coefficients. Then the value of 11430X is ____________.

Moderate
Solution

X=r=110r 10Cr2=r=110r10Cr10Cr=10r=1109Cr110C10r=10.cofficient of x9 in (1+x)9(1+x)10=10.cofficient in(1+x)19=1019C9

Now, X1430=1019C91430= 19C9143= 19C911×13

                   =19×17×168=19×34=646



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