Understanding Permutations in Mathematics

$$n! = n \cdot (n-1) \cdot (n-2) \cdot \dots \cdot 1$$

For example:

• $5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120$
• $3! = 3 \cdot 2 \cdot 1 = 6$

By convention, $0!$ is defined as $1$. This might seem counterintuitive, but it is mathematically necessary to ensure that our permutation formulas work correctly when we select all available items from a set.

Permutations of a Whole Set

When you are arranging $n$ distinct objects into $n$ available positions, the number of possible permutations is simply $n!$.

Worked Example 1: How many ways can 5 students stand in a single line for a photograph? Since there are 5 students and 5 positions, we calculate:

$$5! = 120$$

There are 120 different ways the students can be arranged.

Permutations of a Subset ($nPr$)

In many real-world scenarios, we do not use all the items available to us. We might have 10 athletes in a race, but we only care about the number of ways the Gold, Silver, and Bronze medals can be awarded. This is a permutation of $r$ items taken from a set of $n$ items.

The formula for calculating these permutations is:

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

Where:

• $n$ is the total number of items in the set.
• $r$ is the number of items being arranged.

Worked Example 2: Using our race example: There are 10 athletes ($n=10$), and we want to find the arrangements for the top 3 positions ($r=3$). Using the formula:

$$P(10, 3) = \frac{10!}{(10-3)!} = \frac{10!}{7!}$$

We can expand the factorials to simplify:

$$\frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 10 \cdot 9 \cdot 8 = 720$$

There are 720 different ways the medals can be awarded.

Permutations with Repetition (Indistinguishable Objects)

A more advanced concept arises when some items in your set are identical. If you are arranging the letters in the word "APPLE," you will notice that there are two 'P's. Swapping the first 'P' with the second 'P' does not create a visually different arrangement. To account for this, we must divide the total permutations by the factorials of the counts of each repeated item.

The formula for permutations of $n$ objects where $n_1$ are of one type, $n_2$ of another, and so on, is:

$$\frac{n!}{n_1! \cdot n_2! \cdot \dots \cdot n_k!}$$

Worked Example 3: How many distinct ways can you arrange the letters in "BANANA"?

1. Total letters ($n$): 6
2. Count of 'B': 1
3. Count of 'A': 3
4. Count of 'N': 2

Applying the formula:

$$\frac{6!}{1! \cdot 3! \cdot 2!} = \frac{720}{1 \cdot 6 \cdot 2} = \frac{720}{12} = 60$$

There are 60 unique arrangements of the letters in "BANANA."

Summary Table

To help you choose the right method, refer to this quick guide:

By mastering these three variations, you can solve the vast majority of arrangement-based problems encountered in mathematics and probability.