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If sum(r=0)^25 {.^50Cr*^(50-r)C(25-r)}=K...

If `sum_(r=0)^25 {.^50C_r*^(50-r)C_(25-r)}=K(.^50C_25),` then `K` is equal to

A

`2^(24)`

B

`2^(25)-1`

C

`(25)^(2)`

D

`2^(25)`

Text Solution

Verified by Experts

The correct Answer is:
D

`sum_(r=0)^(25)""^(50)C_(r).""^((50-r))C_(25-r)`
`sum_(r=0)^(25) (50!)/(r!(50-r)!)xx((50-r)!)/(25! xx (25-r)!)`
Now multiply and divide by 25!
`=sum_(r=0)^(25) (50!)/(r!) xx (25!)/(25!^(2).(25-r)!) = sum_(r=0)^(25) (50!)/(25! .25!)xx (25!)/(r!(25-r)!)=sum_(r=0)^(25)""^(50)C_(25).""^(25)C_(r)`
`""^(50)C_(25)sum_(r=0)^(25) ""^(25)C_(r)= ""^(50)C_(25).2^(25)=""^(50)C_(25).k`
Hence `k=2^(25)`
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