If frac{{}^{n+1}C_{r+1}}{{}^{n+1}C_r} = frac{ | vedclass

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(C) We are given the expression $\frac{{}^{n+1}C_{r+1}}{{}^{n+1}C_r} = \frac{n-r+1}{m}$.Using the formula ${}^{n}C_{r} = \frac{n!}{r!(n-r)!}$,we have:

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$r+1$

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(C) We are given the expression $\frac{{}^{n+1}C_{r+1}}{{}^{n+1}C_r} = \frac{n-r+1}{m}$.Using the formula ${}^{n}C_{r} = \frac{n!}{r!(n-r)!}$,we have:$\frac{{}^{n+1}C_{r+1}}{{}^{n+1}C_r} = \frac{\frac{(n+1)!}{(r+1)!(n-r)!}}{\frac{(n+1)!}{r!(n-r+1)!}} = \frac{r!(n-r+1)!}{(r+1)!(n-r)!}$.Simplifying the factorials:$= \frac{r! 	imes (n-r+1) 	imes (n-r)!}{(r+1) 	imes r! 	imes (n-r)!} = \frac{n-r+1}{r+1}$.Comparing this with the given expression $\frac{n-r+1}{m}$,we get $m = r+1$.Thus,option $C$ is correct.

If $\frac{{}^{n+1}C_{r+1}}{{}^{n+1}C_r} = \frac{n-r+1}{m}$,then $m =$

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