Number of Permutations Without Repetition
Category : JEE Main & Advanced
(1) Arranging \[n\] objects, taken \[r\] at a time equivalent to filling \[r\] places from \[n\] things.
The number of ways of arranging = The number of ways of filling \[r\] places.
\[=n(n-1)\,(n-2).......(n-r+1)\]
\[=\frac{n(n-1)\,(n-2).....(n-r+1)((n-r)!)}{(n-r)!}=\frac{n\,!}{(n-r)!}{{=}^{n}}{{P}_{r}}\]
(2) The number of arrangements of \[n\] different objects taken all at a time \[={{\,}^{n}}{{P}_{n}}=n!\]
(i) \[^{n}{{P}_{0}}=\frac{n\,!}{n\,!}=1;{{\,}^{n}}{{P}_{r}}=n\,{{.}^{n-1}}{{P}_{r-1}}\]
(ii) \[0\,!=1;\,\,\frac{1}{(-r)\,!}=0\] or \[(-r)\,!=\infty \,\,\,(r\in N)\]
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