For 1 leq r leq n,the value of frac{1}{r+1}le | vedclass

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(B) Given expression: $\frac{1}{r+1}\left\{{ }^n P_{r+1}-{ }^{(n-1)} P_{r+1}\right\}$ Using the formula ${ }^n P_r = \frac{n!}{(n-r)!}$,we have: $= \f

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${ }^{n-1} P_r$

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(B) Given expression: $\frac{1}{r+1}\left\{{ }^n P_{r+1}-{ }^{(n-1)} P_{r+1}\right\}$ Using the formula ${ }^n P_r = \frac{n!}{(n-r)!}$,we have: $= \frac{1}{r+1} \left[ \frac{n!}{(n-(r+1))!} - \frac{(n-1)!}{(n-1-(r+1))!} \right]$ $= \frac{1}{r+1} \left[ \frac{n!}{(n-r-1)!} - \frac{(n-1)!}{(n-r-2)!} \right]$ $= \frac{1}{r+1} \left[ \frac{n(n-1)!}{(n-r-1)!} - \frac{(n-1)!(n-r-1)}{(n-r-1)!} \right]$ $= \frac{(n-1)!}{(r+1)(n-r-1)!} [n - (n-r-1)]$ $= \frac{(n-1)!}{(r+1)(n-r-1)!} [r+1]$ $= \frac{(n-1)!}{(n-r-1)!} = { }^{n-1} P_r$

For $1 \leq r \leq n$,the value of $\frac{1}{r+1}\left\{{ }^n P_{r+1}-{ }^{(n-1)} P_{r+1}\right\}$ is equal to

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