Permutation of n things taking r at a time P(n,r)

Permutation is a fundamental concept in combinatorics that deals with the arrangement of objects in a specific order. When we talk about the permutation of n things taking r at a time, denoted as P(n, r), we are interested in the number of ways to arrange r objects out of n distinct objects.

Understanding Permutations

Before diving into the formula and examples, let's clarify what we mean by permutations:

Permutation: An ordered arrangement of objects.
n: The total number of distinct objects.
r: The number of objects to be arranged at a time.

Formula for P(n, r)

The formula for calculating the number of permutations of n things taken r at a time is given by:

$$ P(n, r) = \frac{n!}{(n-r)!} $$

where n! (n factorial) is the product of all positive integers from 1 to n.

Important Points

Here are some important points to remember about permutations:

The order of arrangement is important in permutations.
If r = n, then P(n, n) = n!, which means we are arranging all objects.
If r > n, then P(n, r) = 0, since we cannot arrange more objects than we have.
P(n, 0) = 1, since there is only one way to arrange zero objects.

Differences and Important Points

Aspect	Permutation (P(n, r))	Combination (C(n, r))
Order	Order of selection matters	Order of selection does not matter
Formula	$P(n, r) = \frac{n!}{(n-r)!}$	$C(n, r) = \frac{n!}{r!(n-r)!}$
Example	Arranging books on a shelf	Choosing books to take home from a shelf
When to use	When the arrangement or sequence is important	When only the selection is important, not the order

Examples

Let's look at some examples to better understand P(n, r):

Example 1

Question: How many ways can 3 medals (gold, silver, bronze) be awarded to 8 athletes?

Solution:

Here, n = 8 (athletes) and r = 3 (medals). We want to find P(8, 3).

$$ P(8, 3) = \frac{8!}{(8-3)!} = \frac{8!}{5!} = \frac{8 \times 7 \times 6 \times 5!}{5!} = 8 \times 7 \times 6 = 336 $$

There are 336 different ways to award the 3 medals to the athletes.

Example 2

Question: In how many ways can 4 students be seated in a row of 6 chairs?

Solution:

Here, n = 6 (chairs) and r = 4 (students). We want to find P(6, 4).

$$ P(6, 4) = \frac{6!}{(6-4)!} = \frac{6!}{2!} = \frac{6 \times 5 \times 4 \times 3 \times 2!}{2!} = 6 \times 5 \times 4 \times 3 = 360 $$

There are 360 different ways to seat the 4 students in the row of 6 chairs.

Example 3

Question: If there are 10 different books, in how many ways can 5 be selected and arranged on a shelf?

Solution:

Here, n = 10 (books) and r = 5 (books to arrange). We want to find P(10, 5).

$$ P(10, 5) = \frac{10!}{(10-5)!} = \frac{10!}{5!} = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{5!} = 10 \times 9 \times 8 \times 7 \times 6 = 30,240 $$

There are 30,240 different ways to select and arrange 5 books on a shelf from a selection of 10 books.

Practice Problems

Find P(7, 3).
How many 4-letter words (real or nonsense) can be formed from the English alphabet when letters can be repeated?
In how many ways can a president, vice-president, and treasurer be selected from a club with 12 members?

Understanding permutations and how to calculate them is essential for solving a variety of problems in mathematics and related fields. Remember to consider the order of arrangement and the specific conditions of the problem when calculating permutations.