Question

The total number of permutations of \(\mathrm{n}(\mathrm{n}>1\) ) different things taken not more than \(\mathrm{r}\) at a time, when each things may be repeated any number of times is (a) \(\left[\left\\{\mathrm{n}\left(\mathrm{n}^{\mathrm{n}}-1\right)\right\\} /\\{\mathrm{n}-1\\}\right]\) (b) \(\left[\left\\{\left(\mathrm{n}^{\mathrm{r}}-1\right)\right\\} /\\{\mathrm{n}-1\\}\right]\) (c) \(\left[\left\\{n\left(n^{r}-1\right)\right\\} /\\{n-1\\}\right]\) (d) \([\\{n(n-r)\\} /\\{n-1\\}]\)

Short answer

The short answer is: (c) \(\frac{n\left(n^r-1\right)}{n-1}\).

Step-by-step solution

A permutation is an arrangement of n objects in a specific order. When repetition is allowed, each of the n objects may appear any number of times within the permutation. The total number of permutations when repetition is allowed can be calculated as: \(n^r\), where n is the number of objects, and r is the number of times those objects are arranged. #Step 2: Considering permutations with not more than r at a time#

The given question asks for permutations of n things taken not more than r at a time. This means that we need to sum up the number of permutations for each i, where 1 ≤ i ≤ r. So, we need to calculate the sum of the series: \(n^1 + n^2 + \cdots + n^r\). #Step 3: Calculate the sum of the series#

To find the sum of a geometric series, we can use the formula: \(\frac{a_{1}\left(1-r^n\right)}{1-r}\), where \(a_{1}\) is the first term of the series, r is the common ratio, and n is the number of terms in the series. In our problem, the first term, \(a_{1}\), is equal to \(n^1 = n\), and the common ratio is equal to n. To find the sum of the series, we will plug these values into the formula: \(S = \frac{n\left(1-n^r\right)}{1-n}\). #Step 4: Match the expression for the sum of the series with the given options#

We now need to match our calculated expression for the sum of the series (number of permutations) with one of the provided options in the question. (a) \(\frac{n\left(n^n-1\right)}{n-1}\) \\ (b) \(\frac{n^r-1}{n-1}\) \\ (c) \(\frac{n\left(n^r-1\right)}{n-1}\) \\ (d) \(\frac{n(n-r)}{n-1}\) Comparing the expressions, we find that the correct answer is (c) \(\frac{n\left(n^r-1\right)}{n-1}\).

Key concepts

Permutation Formula

Permutations are central to understanding the various ways in which objects can be arranged. The basic permutation formula tells us how many different ways we can arrange a set of objects in a particular order. In situations where repetition is allowed, and we're dealing with not just a single arrangement but rather multiple arrangements up to a certain length, the formula adapts.

For instance, consider you have a set of n unique items and you want to find out all possible arrangements taking r items at a time with repetition, the total permutations is given by the simple exponentiation of n raised to the power of r, expressed as \(n^r\). To put it more contextually, if you have 3 different colored beads and you want to create necklaces of length 2 allowing colors to repeat, you have \(3^2\) or 9 possible combinations.

It's like rolling a die (where each face represents an object): if you could roll it r times (the number of objects you pick each time), the permutation formula would account for all the possible outcomes of those r rolls.

Geometric Series Sum

Now, imagine that instead of one necklace, you decided to create necklaces of all possible lengths up to r. This situation generates a geometric series because the number of permutations for each possible length forms a sequence where each term is multiplied by a common ratio to obtain the next. The sum of such a series is crucial in understanding the permutations with repetition for varying lengths.

Geometric series can be summed up beautifully using a mathematical formula. Given a starting term \(a_1\), and a ratio \(r\), the sum of the first n terms \(S_n\) can be found using the formula \(S_n = \frac{a_{1}(1-r^n)}{1-r}\). This succinct equation collapses what could be a very large sum into a straightforward calculation. Taking the example of a geometric series \(n + n^2 + n^3 + ... + n^r\), where the ratio between consecutive terms is \(n\), we use the formula to find that the sum would be \( \frac{n(1-n^r)}{1-n}\).

It's like adding up all the potential outcomes from creating 1-bead, 2-bead, all the way up to r-bead necklaces sequentially, considering the number of permutations for each length.

Arrangements in Mathematics

In mathematics, an arrangement refers to the order or sequence of elements within a set. When we talk about arrangements with repetition, we relax the rule that typically prohibits elements from appearing more than once. This opens up the fascinating world of permutations to an even greater array of possibilities.

Whether you're solving a puzzle, forming a line of people where twins are allowed to appear multiple times, or selecting toppings for your pizza irrespective of their prior use, you're dealing with the concept of arrangements in mathematics with repetition. The formulas and the concepts of geometric series directly apply to these scenarios giving us a splendid toolkit to systematically compute possible combinations from an otherwise mind-boggling array of choices.