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The value of sum(r=0)^40 rC(40,r) C(30,r...

The value of `sum_(r=0)^40 rC(40,r) C(30,r)`

A

`40.^(69)C_(29)`

B

`40.^(70)C_(30)`

C

`"^(69)C_(29)`

D

`"^(70)C_(30)`

Text Solution

Verified by Experts

The correct Answer is:
A

`sum_(r =0)^(40) r.(40)/(r).^(39) C_(r-1).^(39)C_(r)`
`= 40 sum_(r =1)^(40) "^(39)C_(r-1).^(30)C_(r)`
`(1+x)^(39) = ^(39)C_(0) + ^(39)C_(1)x+ .....+ ^(39)C_(r)x^(r)+ ...... + ^(39)C_(39)x^(39).....(i)`
`(x+1)^(30) = ^(30)C_(0)x ^(30) + ^(30)C_(1)x^(29) + ^(30)C_(2)x^(28) + ..... + ^(30)C_(30)....(ii)`
`(1+x)^(39) = (x+1)^(30) = [ "^(30)C_(0). ^(30)C_(1). ^(30)C_(2) + ...... + ^(39)C_(29).^(30)C_(30)]`
Multiple (i) by (ii) and comparing the coeff. of `x^(29)`.
= `40.^(60)C_(29)`
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