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Question

The value of
1 20C0+2 20C1 30C1+1 20C0++3 20C2 30C2++21 20C20 30C20 is 50C20+20 49Cq, (where q is an even number), then qq+5 is

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Solution

20r=0(r+1) 20Cr 30Cr=20r=0 20Cr 30Cr+20r=1 r20r 19Cr1 30Cr

(1+x)20= 20C0+ 20C1 x+ 20C2 x2+ ...[1](1+x)30= 30C0 x30+ 30C1 x29+ 30C2 x28+ ...[2]

Coefficient of x30 in the product of [1] and [2] will be
=20r=0 20Cr 30Cr

Coefficient of x30 in (1+x)50= 50C30
20r=0 20Cr 30Cr= 50C30

(1+x)19= 19C0+ 19C1 x+ 19C2 x2+ ...[3](1+x)30= 30C0 x30+ 30C1 x29+ 30C2 x28+ ...[4]

Coefficient of x29 in the product of [3] and [4] will be
=20r=1 19Cr1 30Cr

Coefficient of x29 in (1+x)49= 49C29= 49C20
20r=1 19Cr1 30Cr= 49C20

20r=0(r+1) 20Cr 30Cr= 50C20+20 49C20

q=20

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