Question

The total number of permutation of n different things taken not more than r at a time, when each thing may be repeated any number of times is$\left( \text{a} \right)\text{ }{{n}^{r}}-1$$\left( \text{b} \right)\text{ }\dfrac{n\left( {{n}^{r}}-1 \right)}{n-1}$$\left( \text{c} \right)\text{ }\dfrac{{{n}^{r}}-1}{n-1}$$\left( \text{d} \right)\text{ }\dfrac{n\left( {{n}^{r}}-1 \right)}{n+1}$

Hint: To solve the given question, we will make use of the fact that the minimum number of things which are taken out is 1 and the maximum number of things taken at a time is r. Here, we will assume that x things are taken out from n things such that $1\le x\le r.$ Then, we will make use of the fact that if we have to choose x things from n things and there are as many repetitions of these things as we want, then the number of ways of doing this will be ${{n}^{x}}.$ Then, we will put different values of x from 1 to r and then we will add all these. On adding, we will get a series whose sum will be found out by using the appropriate formula.

Complete step by step solution:
To start with, we are given that we cannot take more than r things out of n at a time i.e. the maximum number of things which can be taken out at a time is r. Let us assume that we have taken out x things out of n things such that $1\le x\le r.$ Now, we are given that when we take one thing out, the number of distinct things remains the same, i.e. n. This means that things have a sufficient number of repetitions. Thus, when we will take one thing, next time, we will still have n distinct things. Now, we know that x varies from 1 to r. Let x = 1, i.e. we take out one thing at a time. The number of ways of doing this is n.
Now, we assume that x = 2. This means that we have to select two things at a time. The number of ways of selecting the first thing is n. Now, we know that there are enough repetitions of things such that even after taking one thing, the number of distinct things remains n. Now, the number of ways of selecting the second thing is n. Thus, the total number of permutations become $n\times n={{n}^{2}}.$
Similarly, when n = 3, the total number of ways of selecting the first, second, and third things are n, n, and n respectively. Thus, the total number of permutations become $n\times n\times n={{n}^{3}}.$ Thus, when x = r, the total number of permutations will be ${{n}^{r}}.$
Thus, the overall number of permutations will be the sum of all these cases. Thus,
$\text{Total permutations}=n+{{n}^{2}}+{{n}^{3}}+......+{{n}^{r}}$
We can see that the above series is a geometric progression with common ratio n. We know that the sum of the terms of a GP is equal to $a\left( \dfrac{1-{{p}^{q}}}{1-p} \right)$ where a is the first term, p is the common ratio and q are the total number of terms. Thus, we will get,
$\Rightarrow \text{Total permutation}=\dfrac{n\left( 1-{{n}^{r}} \right)}{1-n}$
$\Rightarrow \text{Total permutation}=\dfrac{n\left( {{n}^{r}}-1 \right)}{n-1}$
Hence, option (b) is the right answer.

Note: In the question, it is given that each thing may be repeated at any number of times. In our case, each thing should be repeated at least r times because when we have selected r things at a time, then it is possible that we have selected each thing of the same kind.