^nC_r div ^nC_{r-1} = | vedclass

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Question

(C) The formula for combinations is given by $^nC_r = \frac{n!}{r!(n-r)!}$.Now,consider the ratio:$\frac{^nC_r}{^nC_{r-1}} = \frac{\frac{n!}{r!(n-r)!}

Vedclass · Accepted Answer

$\frac{n-r+1}{r}$

Vedclass · Answer

(C) The formula for combinations is given by $^nC_r = \frac{n!}{r!(n-r)!}$.Now,consider the ratio:$\frac{^nC_r}{^nC_{r-1}} = \frac{\frac{n!}{r!(n-r)!}}{\frac{n!}{(r-1)!(n-(r-1))!}}$$= \frac{n!}{r!(n-r)!} 	imes \frac{(r-1)!(n-r+1)!}{n!}$$= \frac{(r-1)!}{r!} 	imes \frac{(n-r+1)!}{(n-r)!}$Since $r! = r 	imes (r-1)!$ and $(n-r+1)! = (n-r+1) 	imes (n-r)!$,we get:$= \frac{(r-1)!}{r(r-1)!} 	imes \frac{(n-r+1)(n-r)!}{(n-r)!}$$= \frac{n-r+1}{r}$Thus,the correct option is $C$.

$^nC_r \div ^nC_{r-1} = $

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