Question

The total number of permutations of $n\left( {n > 1} \right)$ different things have taken not more than $r$ at a time, when each thing may be repeated any number of times?
$\eqalign{
  & 1.\dfrac{{n\left( {{n^r} - 1} \right)}}{{\left( {n - 1} \right)}} \cr
  & 2.\dfrac{{\left( {{n^r} - 1} \right)}}{{\left( {n - 1} \right)}} \cr
  & 3.\dfrac{{n\left( {{r^n} - 1} \right)}}{{\left( {n - 1} \right)}} \cr
  & 4.\dfrac{{n\left( {n - r} \right)}}{{\left( {n - 1} \right)}} \cr} $

Answer 1

Hint: This question will be in the form of mathematical induction. We need to find out the pattern for how the given permutations can be arranged. Therefore, we start by taking values for $n$ starting from $1$. After we find out the pattern, we can reach the final answer.

Complete step-by-step solution:
Let us consider that the number of ways for arranging a thing is $n$
Similarly, the number of ways to arrange a thing twice will be
$\eqalign{
  & \Rightarrow n \times n = {n^2} \cr
  & \cr} $
The number of ways keeps on increasing.
$n$will be ${n^2},{n^3},{n^4}....{n^r}$
Where, $r$is the last term in the sequence.
$n + {n^2} + {n^3} + {n^4}....{n^r} = \dfrac{{n({n^r} - 1)}}{{n - 1}}$
Therefore, the final answer is $\dfrac{{n({n^r} - 1)}}{{n - 1}}$
Hence, option (1) is the correct answer for the given question.

Additional Information:
Mathematical induction is a mathematical proof technique. It is usually used to prove a given statement. Permutation means arranging a given set of numbers or characters in a particular order. The formula used for finding a permutation is ${}^n{P_r} = \dfrac{{n!}}{{(n - r)!}}$ where, ${}^n{P_r}$ is the permutation, $n$is the total number of objects and $r$ is the number of objects that are selected to find out the permutation.

Note: The question just mentions permutation, but as we read it, we understand that it does not ask for the permutations. It has to be solved by finding the pattern rather than arranging the terms. Also, the options given look similar, be careful while choosing the right answer.