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If ^n+1C(r+1)dot^n Crdot^(n-1)C(r-1)=11 ...

If `^n+1C_(r+1)dot^n C_rdot^(n-1)C_(r-1)=11 :6:3,` then `n r=` `20` b. `30` c. `40` d. `50`

A

`20`

B

`30`

C

`40`

D

`50`

Text Solution

Verified by Experts

The correct Answer is:
D
Given,
`(.^(n+1)C_(r+1))/(.^(n)C_(r))= 11/6` or `((n+1)/(r+1)xx.^(n)C_(r))/(.^(n)C_(r)) = 11/6`
or `6n+6 = 11 r + 11` or `6n-11r = 5" "(1)`
Also,
`(.^(n)C_(r))/(.^(n-1)C_(r-1)) = 6/3` or `(n/rxx.^(n-1)C_(r-1))/(.^(n-1)C_(r-1)) = 6/3` or `n = 2r " " (2)`
From (1) and (2), `r = 5` and `n = 10`,
`:. nr = 50`
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