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The total number of permutations of $n$ different things taken not more than $r$ at a time, where each thing may be repeated any number of times, is

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\frac{n (n^{n} - 1)}{n - 1}

\frac{(n^{r} - 1)}{n - 1}

\frac{n (n^{r} - 1)}{n - 1}

\frac{(n^{n} - 1)}{n - 1}

answer is C.

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Detailed Solution

Let

x

number of things are taken from

n

number of things and

x

ranges values from

1

r

.
Therefore,
For,

x = 1

, the number of things taken out

= n^{1} = n

.
For,

x = 2

, the number of things taken out

= n^{2}

.
For,

x = 3

, the number of things taken out

= n^{3}

.
And so on.
Now,
For,

x = r

, the number of things taken out

= n^{r}

.
So,
The total number of permutations

=

Sum of all the cases above.

= n + n^{2} + n^{3} + . . . . . . . + n^{r}

.
To find the value of the above expression, we will be using the formula of geometric progression, as the above expression is in the form of geometric progression.
Formula is given as:

\Rightarrow

Sum of

n

terms of a geometric progression

= \frac{a (1 - n^{r})}{(1 - n)}

Where,

a

is the first term of the expression,

p

is the common ratio and

q

is the total number of terms.
Now, according to the above expression,

\Rightarrow a = n

\Rightarrow p = \frac{n^{2}}{n}

\Rightarrow p = n

And,

\Rightarrow q = r

Therefore, putting all the values in the given formula,
Total number of permutations

= \frac{n (1 - n^{r})}{(1 - n)}

= \frac{- n (n^{r} - 1)}{- (n - 1)}

Therefore, canceling both the minus signs, we get our required answer.

\Rightarrow

Total number of permutations

= \frac{n (n^{r} - 1)}{(n - 1)}

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The total number of permutations of n different things taken not more than r at a time, where each thing may be repeated any number of times, is

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The total number of permutations of $n$ different things taken not more than $r$ at a time, where each thing may be repeated any number of times, is