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The total number of permutations of \mathrm{n(>1)}  different things taken not more that \mathrm{r} at a time, when a thing may be repeated any number of times, is

Option: 1

\mathrm{\frac{n}{n-1}\left(n^{r}-1\right)}


Option: 2

\mathrm{\frac{n^{r-1}}{n-1}}


Option: 3

\mathrm{\frac{n^{\prime}+1}{n+1}}


Option: 4

\mathrm{\frac{n^{r}+n}{n-1}}


Answers (1)

best_answer

When \mathrm{k(1 \leq k \leq r)} things are arranged, the number of possible arrangements, when repetition is allowed is

\mathrm{\underbrace{n \times n \times \ldots \ldots \times n}_{k \text { times }}=n^{k}}

Thus, the total possible arrangements is

\mathrm{n+n^{2}+n^{3}+\ldots+n^{r}=\frac{n\left(n^{r}-1\right)}{n-1}}.
 

Posted by

seema garhwal

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