A) \[^{n-1}{{P}_{r}}+r{{\,}^{n-1}}{{P}_{r-1}}\]
B) \[n.{{\ }^{n-1}}{{P}_{r}}{{+}^{n-1}}{{P}_{r-1}}\]
C) \[n{{(}^{n-1}}{{P}_{r}}{{+}^{n-1}}{{P}_{r-1}})\]
D) \[^{n-1}{{P}_{r-1}}{{+}^{n-1}}{{P}_{r}}\]
Correct Answer: A
Solution :
\[^{n-1}{{P}_{r}}+r{{.}^{n-1}}{{P}_{r-1}}\] \[=\frac{(n-1)\,!}{(n-1-r)\,!}+r\frac{(n-1)\,!}{(n-r)\,!}\] \[\left( \because \,\,{{\,}^{n}}{{P}_{r}}=\frac{n\,!}{(n-r)\,!} \right)\] = \[\frac{(n-1)\,!}{(n-1-r)\,!}\,\,\left\{ 1+r.\frac{1}{n-r} \right\}\] = \[\frac{(n-1)\,!}{(n-1-r)\,!(n-r)\,!}\left( \frac{n}{n-r} \right)=\frac{n\,!}{(n-r)\,!}={{\,}^{n}}{{P}_{r}}\]. Aliter: We know that \[^{n-1}{{C}_{r}}+{{\,}^{n-1}}{{C}_{r-1}}={{\,}^{n}}{{C}_{r}}\] Þ \[\frac{^{n-1}{{P}_{r}}}{r\,!}+\frac{^{n-1}{{P}_{r-1}}}{(r-1)\,!}=\frac{^{n}{{P}_{r}}}{r\,!}\] Þ \[^{n-1}{{P}_{r}}+r\,.{{\,}^{n-1}}{{P}_{r-1}}={{\,}^{n}}{{P}_{r}}\].
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