Permutations and Combinations

In the counting principles lesson, you learned the fundamental counting principle and noticed that sometimes different orderings give the same outcome. Permutations and combinations formalize the distinction: permutations count arrangements where order matters, and combinations count selections where order does not matter. These formulas appear throughout probability, statistics, and discrete mathematics.

Permutations (Order Matters)

A permutation is an arrangement of objects in a specific order. The number of ways to arrange $r$ objects chosen from $n$ distinct objects is:

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

Why this formula works: The first position has $n$ choices, the second has $n - 1$, and so on down to $n - r + 1$ choices. This product equals $\frac{n!}{(n-r)!}$.

Example 1: Race Finishers

In a race with 10 runners, how many ways can the gold, silver, and bronze medals be awarded?

$P(10, 3) = \frac{10!}{7!} = 10 \times 9 \times 8 = 720$

Example 2: Arranging Letters

How many different 4-letter arrangements can be made from the letters in PRIME (5 letters, no repeats)?

$P(5, 4) = \frac{5!}{1!} = 120$

Special Case: Arranging All Objects

When $r = n$ (you arrange all $n$ objects):

$P(n, n) = \frac{n!}{0!} = n!$

This is why the number of ways to arrange 6 books on a shelf is $6! = 720$.

Combinations (Order Does Not Matter)

A combination is a selection of objects where order is irrelevant. The number of ways to choose $r$ objects from $n$ distinct objects is:

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

The notation $\binom{n}{r}$ is read “n choose r” and is the same binomial coefficient from the binomial theorem.

Why divide by $r!$? A permutation counts each group $r!$ times (once for each ordering). Dividing removes the duplicate orderings:

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

Example 3: Choosing a Committee

From a club of 12 members, how many ways can a 4-person committee be chosen?

$C(12, 4) = \frac{12!}{4! \cdot 8!} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = \frac{11{,}880}{24} = 495$

Example 4: Lottery Numbers

A lottery draws 6 numbers from 1 to 49. How many possible outcomes?

$C(49, 6) = \frac{49!}{6! \cdot 43!} = \frac{49 \times 48 \times 47 \times 46 \times 45 \times 44}{720} = 13{,}983{,}816$

This is why lotteries are hard to win — nearly 14 million possible combinations.

How to Tell: Permutation or Combination?

Ask: Does the order of selection matter?

Key words that signal order matters: arrange, rank, assign roles, first/second/third Key words that signal order does not matter: choose, select, group, committee, team, hand (of cards)

Permutations with Repetition

When some objects are identical, not all rearrangements produce distinct outcomes. If you have $n$ objects where $n_1$ are identical of type 1, $n_2$ of type 2, etc.:

$\text{Distinct permutations} = \frac{n!}{n_1! \cdot n_2! \cdot \cdots \cdot n_k!}$

Example 5: Rearranging MISSISSIPPI

MISSISSIPPI has 11 letters: M(1), I(4), S(4), P(2).

$\frac{11!}{1! \cdot 4! \cdot 4! \cdot 2!} = \frac{39{,}916{,}800}{1 \times 24 \times 24 \times 2} = \frac{39{,}916{,}800}{1{,}152} = 34{,}650$

Circular Permutations (Brief)

When arranging $n$ objects in a circle, rotations of the same arrangement are considered identical. Fix one object’s position and arrange the rest:

$\text{Circular permutations} = (n - 1)!$

Example: 8 people around a circular table: $(8-1)! = 7! = 5{,}040$ arrangements.

Combining Counting Techniques

Many problems require using permutations, combinations, and the counting principle together.

Example 6: Mixed Committee

From 7 men and 5 women, form a committee of 3 men and 2 women.

Choose 3 men from 7: $C(7, 3) = 35$

Choose 2 women from 5: $C(5, 2) = 10$

By the counting principle: $35 \times 10 = 350$ committees.

Example 7: At Least One Condition

From a standard 52-card deck, how many 5-card hands contain at least one ace?

Complement method: Total hands minus hands with no aces.

Total 5-card hands: $C(52, 5) = 2{,}598{,}960$

Hands with no aces (choose 5 from 48 non-aces): $C(48, 5) = 1{,}712{,}304$

Hands with at least one ace: $2{,}598{,}960 - 1{,}712{,}304 = 886{,}656$

Real-World Application: Product Testing in Retail

A retail company has 20 new products and wants to select 5 for a trial display. Additionally, the 5 selected products must be arranged in a specific order on the shelf (position 1 through 5).

Step 1 — How many ways to choose which 5 products?

$C(20, 5) = \frac{20!}{5! \cdot 15!} = 15{,}504$

Step 2 — How many ways to arrange 5 products on the shelf?

$5! = 120$

Total arrangements: $15{,}504 \times 120 = 1{,}860{,}480$

Or equivalently: $P(20, 5) = \frac{20!}{15!} = 20 \times 19 \times 18 \times 17 \times 16 = 1{,}860{,}480$.

This shows that $P(n, r) = C(n, r) \times r!$ — choosing and then ordering is the same as permuting directly.

Common Mistakes

1. Using permutations when combinations are needed (and vice versa). Always ask: does order matter?
2. Forgetting to divide by $r!$ for combinations. If you get a suspiciously large answer for a “choose a team” problem, you probably computed a permutation.
3. Not handling identical objects. “How many arrangements of BANANA?” requires dividing by the repeated-letter factorials.
4. Adding when you should multiply. Choosing 3 men AND 2 women uses multiplication. Choosing 3 men OR 2 women (one or the other, not both) uses addition.

Practice Problems

Key Takeaways

• Permutations count ordered arrangements: $P(n, r) = \frac{n!}{(n-r)!}$
• Combinations count unordered selections: $C(n, r) = \frac{n!}{r!(n-r)!}$
• The key question is always: does order matter?
• For identical objects, divide by the factorials of the repeat counts
• Circular permutations fix one position: $(n-1)!$ arrangements
• Combine formulas using the counting principle (multiply across independent choices, add across exclusive categories)
• $P(n, r) = C(n, r) \times r!$ — permuting = choosing then ordering

Return to College Algebra for more topics in this section.