Learn how to solve mathematics word problems containing Permutations and Combinations using formulas.

- Permutations: The number of ways to choose a sample of k elements from a set of n distinct objects where order does matter, and replacements are not allowed. For a permutation problem, use this formula:

\(\color{blue}{_{n}P_{k }= \frac{n!}{(n-k)!}}\) - Combination: The number of ways to choose a sample of r elements from a set of n distinct objects where order does not matter, and replacements are not allowed. For a combination problem, use this formula:

\(\color{blue}{_{ n}C_{r }= \frac{n!}{r! (n-r)!}}\) - Factorials are products, indicated by an exclamation mark. For example, \(4!\) Equals: \(4×3×2×1\). Remember that \(0!\) is defined to be equal to \(1\).

### Example 1:

How many ways can first and second place be awarded to \(10\) people?

**Solution:**

Since the order matters, we need to use permutation formula where n is \(10\) and k is \(2\). Then: \(\frac{n!}{(n-k)!}=\frac{10!}{(10-2)!}=\frac{10!}{8!}=\frac{10×9×8!}{8!}\), remove \(8!\) from both sides of the fraction. Then: \(\frac{10×9×8!}{8!}=10×9=90\)

### Example 2:

How many ways can we pick a team of \(3\) people from a group of \(8\)?

**Solution:**

Since the order doesn’t matter, we need to use combination formula where n is \(8\) and \(r\) is \(3\). Then: \(\frac{n!}{r! (n-r)!}=\frac{8!}{3! (8-3)!}=\frac{8!}{3! (5)!}=\frac{8×7×6×5!}{3! (5)!}=\frac{8×7×6}{3×2×1}=\frac{336}{6}=56\)

## Exercises

### Calculate the value of each.

- \(\color{blue}{4!=}\)
- \(\color{blue}{4!×3!=}\)
- \(\color{blue}{5!=}\)
- \(\color{blue}{6!+3!=}\)
- There are 7 horses in a race. In how many different orders can the horses finish?
- In how many ways can 6 people be arranged in a row?

### Download Combinations and Permutations Worksheet

- \(\color{blue}{24}\)
- \(\color{blue}{144}\)
- \(\color{blue}{120}\)
- \(\color{blue}{726}\)
- \(\color{blue}{5,040}\)
- \(\color{blue}{720}\)