Question

The total number of permutations of $ n $ different thing taken not more than $ r $ at a time, where each thing may be repeated any number of times, is
A. $ \dfrac{{n\left( {{n^n} - 1} \right)}}{{n - 1}} $
B. $ \dfrac{{\left( {{n^r} - 1} \right)}}{{n - 1}} $
C. $ \dfrac{{n\left( {{n^r} - 1} \right)}}{{n - 1}} $
D. $ \dfrac{{\left( {{n^n} - 1} \right)}}{{n - 1}} $

Answer 1

Hint: In the question it is given that the maximum number of things taken out is r and the minimum number of things taken out is 1. Assume a variable $ x $ such that $ x $ things are taken out of $ n $ things and this $ x $ can have values between 1 and $ r $ . Now, when we are choosing $ x $ things then there will be repetition in the selection equals the number of things taken then the number of ways by which each thing is taken is $ {n^x} $ . In the same way we can find the values for all values of $ x $ from 1 to $ r $ .

Complete step-by-step answer:
Let us assume we have taken out $ x $ things from the total $ n $ things and $ x $ have the values from 1 to $ r $ .
For $ x = 1 $ , the number of ways of taking 1 thing out is $ {n^1} = n $ .
For $ x = 2 $ , the number of ways of taking 2 things out is $ {n^2} $ .
Similarly, for $ x = 3 $ , the number of ways of taking 3 things out is $ {n^3} $ .
So, we can write a similar expression for all the values of $ x. $
For $ x = r $ , the number of ways of taking $ r $ things out is $ {n^r} $ .
The total number of permutations will be the sum of all the cases.
 $ {\rm{textTotal permutations}} = n + {n^2} + {n^3} + ........... + {n^r} $
The above expression is in the form of geometric progression and has a common ratio $ n $ .
The sum of all the terms of geometric progression can be expressed as:
 $ a\left( {\dfrac{{1 - {p^q}}}{{1 - p}}} \right) $
Where, $ a $ is the first term of geometric progression, p is the common ratio geometric progression and q is the total number of terms in the geometric progression.
The first term for the given case is $ n $ .
The last term for the given case is $ {n^r} $ .
The total number of the terms are $ r $ .
 $ \begin{array}{c}
{\rm{\text Total permutations}} = \dfrac{{n\left( {1 - {n^r}} \right)}}{{1 - n}}\\
 = \dfrac{{n\left( {{n^r} - 1} \right)}}{{n - 1}}
\end{array} $
So, the correct answer is “Option C”.

Note: Make sure that each thing in the question can be repeated $ r $ times because on selecting things $ r $ times then only it’s possible that things selected are of the same kind.