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nPr+r(n1)Pr1=?^nP_r+r(n-1)P_{r-1}=?

APrP_r

Bn1Pr1^{n-1}P_{r-1}

CnPr^nP_r

DnPr1^nP_{r-1}

Answer:

nPr^nP_r

Read Explanation:

nPr+r(n1)Pr1^nP_r+r(n-1)P_{r-1}

=(n1)!(nr1)!+r(n1)!(n1r+1)!=\frac{(n-1)!}{(n-r-1)!}+\frac{r(n-1)!}{(n-1-r+1)!}

=(n1)!×[nr(nr)!+r(nr)!]=(n-1)!\times[\frac{n-r}{(n-r)!}+\frac{r}{(n-r)!}]

=(n1)!(nr)!nr+r=\frac{(n-1)!}{(n-r)!}{n-r+r}

=n(n1)!(nr)!=\frac{n(n-1)!}{(n-r)!}

=nPr=^nP_r


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