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\[\left( \text{a} \right)\text{ }{{n}^{r}}-1\]

\[\left( \text{b} \right)\text{ }\dfrac{n\left( {{n}^{r}}-1 \right)}{n-1}\]

\[\left( \text{c} \right)\text{ }\dfrac{{{n}^{r}}-1}{n-1}\]

\[\left( \text{d} \right)\text{ }\dfrac{n\left( {{n}^{r}}-1 \right)}{n+1}\]

Answer
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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.

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