Question

The total no. of permutations of $\mathrm{n}(\mathrm{n}>1)$ different things have taken not more than $\mathrm{r}$ at a time, when each thing may be repeated any number of times is$?$

• A. $\frac{\mathrm{n}\left({\mathrm{n}}^{\mathrm{r}}-1\right)}{\left(\mathrm{n}-1\right)}$
• B. $\frac{\left({\mathrm{n}}^{\mathrm{r}}-1\right)}{(\mathrm{n}-1)}$
• C. $\frac{\mathrm{n}({\mathrm{r}}^{\mathrm{n}}-1)}{(\mathrm{n}-1)}$
• D. $\frac{\mathrm{n}(\mathrm{n}-\mathrm{r})}{(\mathrm{n}-1)}$

Answer 1

The correct option is A

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

Explanation of the correct option:

Total number of things are $\mathrm{n}$.

If we take $1$ thing at a time, then number of ways$=\mathrm{n}$

If we take $2$ thing at a time, then number of ways$={\mathrm{n}}^2$

If we take $3$ thing at a time, then number of ways$={\mathrm{n}}^3$

So, for $\mathrm{r}$ number of things, number of ways $={\mathrm{n}}^{\mathrm{r}}$

Therefore total number of ways from $1$ to $\mathrm{r}$ at a time is given by,

$$\begin{gathered} \mathrm{n}+{\mathrm{n}}^2+{\mathrm{n}}^3+...........+{\mathrm{n}}^{\mathrm{r}} \\ =\mathrm{n}({\mathrm{n}}^{\mathrm{r}-1}+{\mathrm{n}}^{\mathrm{r}-2}+....+1) \\ =\mathrm{n}\frac{({\mathrm{n}}^{\mathrm{r}-1}+{\mathrm{n}}^{\mathrm{r}-2}+....+1)}{(\mathrm{n}-1)}(\mathrm{n}-1) \\ =\mathrm{n}\frac{\left({\mathrm{n}}^{\mathrm{r}}-1\right)}{\left(\mathrm{n}-1\right)} \end{gathered}$$

Therefore, total number of ways is $\mathrm{n}\frac{\left({\mathrm{n}}^{\mathrm{r}}-1\right)}{\left(\mathrm{n}-1\right)}$.

Hence option (A) is the correct option.