Permutations

A permutation is an arrangement of objects where order matters. Picking a president and vice-president, seating people in a row, or setting a lock code are all permutations: swap two people and you get a different result. Permutations are the multiplicative counting principle packed into one handy formula.

Key ideas

What is a permutation?

A permutation of $r$ objects chosen from $n$ different objects is an ordered arrangement of them. For the letters A, B, C, taken two at a time, the permutations are

$$\text{AB, BA, AC, CA, BC, CB}$$

AB and BA count as different, because order matters.

The formula

The number of permutations of $r$ objects chosen from $n$ different objects is

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

Here’s why. There are $n$ choices for the first spot, $n - 1$ for the second, and so on, for $r$ spots:

$$P(n, r) = n \times (n - 1) \times \dots \times (n - r + 1)$$

That’s the first $r$ factors of $n!$, which is exactly $\dfrac{n!}{(n - r)!}$. For example, $P(7, 3) = 7 \times 6 \times 5 = 210$.

Arranging all $n$ objects gives $P(n, n) = \dfrac{n!}{0!} = n!$.

On most calculators, use the nPr key: for $P(7, 3)$, type $7$, then nPr, then $3$.

When to use permutations

Ask: if I swap two of the chosen items, do I get a different result? If yes, order matters, and it’s a permutation. Clues include:

Clue	Example
Different roles or positions	president, vice-president, treasurer
Rankings	first, second, third place
Arrangements in a line	seating, photos, books on a shelf
Codes and “words” with no repeats	a lock code with different digits

If swapping makes no difference (like choosing a committee), see combinations.

Handling restrictions

• Fixed position: place the restricted item first, then arrange the rest. If Ana must sit on the left end of $6$ seats, the other $5$ people fill the remaining seats in $5!$ ways.
• Kept together: glue the items into one block. Arrange the block with the other items, then multiply by the number of ways to arrange the items inside the block.
• Kept apart: use the complement. Count all arrangements, then subtract the ones where the items are together:

$$\text{apart} = \text{total} - \text{together}$$

Worked examples

Example 1: Club executive

A club has $10$ members. In how many ways can it choose a president, a vice-president, and a treasurer?

Solution. The roles are different, so order matters. Choose $3$ from $10$:

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

Example 2: A fixed position

Six people line up for a photo.

• (a) In how many ways can they line up if Ana must be on the left end?
• (b) In how many ways if Ana must be on either end?

Solution. (a) Put Ana on the left end. The other $5$ people fill the other $5$ spots:

$$5! = 120$$

(b) Ana has $2$ choices of end, and then the other $5$ people fill the rest:

$$2 \times 5! = 240$$

Example 3: Together and apart

Seven different books, including two math books, are placed on a shelf.

• (a) In how many ways can they be arranged if the two math books must be side by side?
• (b) In how many ways if the two math books must not be side by side?

Solution. (a) Glue the two math books into one block. Now there are $6$ items to arrange (the block and the $5$ other books): $6!$ ways. Inside the block, the math books can be in either order: $2!$ ways.

$$6! \times 2! = 720 \times 2 = 1440$$

(b) Use the complement. All arrangements: $7! = 5040$.

$$\text{apart} = 5040 - 1440 = 3600$$

Example 4: Making “words”

How many four-letter arrangements can be made from the letters of PLANETS (no letter used twice)? How many of them start with a vowel?

Solution. PLANETS has $7$ different letters. Order matters, so:

$$P(7, 4) = 7 \times 6 \times 5 \times 4 = 840$$

For a vowel first: the vowels are A and E, so the first letter has $2$ choices. The other $3$ spots are filled from the remaining $6$ letters:

$$2 \times P(6, 3) = 2 \times 120 = 240$$

Common mistakes

Using a permutation when order doesn’t matter. Choosing $3$ people for a committee is not a permutation: the same three people in a different order are the same committee. Ask the swap question every time.

Forgetting to arrange inside the block. When items are kept together, the block can be arranged in its own ways. Two books together give a factor of $2!$; three people together give $3!$.

Trying to count “apart” directly. It’s much easier to subtract: total minus together.

Mixing up n and r. In $P(n, r)$, $n$ is how many you have to choose from, and $r$ is how many you arrange. $P(10, 3)$ is $720$, but $P(3, 10)$ doesn’t make sense.

Not filling the restricted spot first. Place the item with a restriction first; otherwise you may count choices that break the rule.

Practice

1. (Warm-up) Evaluate $P(8, 3)$, $P(5, 5)$, and $P(9, 1)$.

Solution

$P(8, 3) = 8 \times 7 \times 6 = 336$. $P(5, 5) = 5! = 120$. $P(9, 1) = 9$.

2. (Warm-up) Does order matter? Say whether each is a permutation.

• (a) awarding gold, silver, and bronze medals to $3$ of $8$ skaters
• (b) choosing $3$ of $8$ students to help at a school event, all doing the same job
• (c) setting a lock code using $4$ different digits

Solution

(a) Yes: swapping gold and silver gives a different result. (b) No: the same three helpers in a different order is the same group. (c) Yes: $1234$ and $4321$ are different codes.

3. (Warm-up) You have $12$ songs. In how many ways can you choose and order the first $4$ songs of a playlist?

Solution

$$P(12, 4) = 12 \times 11 \times 10 \times 9 = 11\,880$$

4. (Core) Eight swimmers are assigned to $8$ lanes.

• (a) How many lane assignments are possible?
• (b) How many if Wei must swim in lane $4$?

Solution

(a) $8! = 40\,320$.

(b) Wei is fixed, so the other $7$ swimmers fill $7$ lanes: $7! = 5040$.

5. (Core) The letters of FRIDAY are arranged.

• (a) How many arrangements are there?
• (b) How many start with F?
• (c) How many have the vowels I and A next to each other?

Solution

FRIDAY has $6$ different letters.

(a) $6! = 720$.

(b) F is fixed first, and the other $5$ letters fill the rest: $5! = 120$.

(c) Glue I and A into a block: $5$ items to arrange, then $2!$ orders inside the block. $5! \times 2! = 240$.

6. (Core) Five friends sit in a row of $5$ seats. Jo and Sam don’t want to sit next to each other. How many seating arrangements are possible?

Solution

Total: $5! = 120$. Together: treat Jo and Sam as a block, giving $4! \times 2! = 48$.

$$\text{apart} = 120 - 48 = 72$$

7. (Core) A bike lock code uses $4$ different digits from $0$ to $9$.

• (a) How many codes are possible?
• (b) How many start with an even digit?

Solution

(a) $P(10, 4) = 10 \times 9 \times 8 \times 7 = 5040$.

(b) The first digit has $5$ choices ($0, 2, 4, 6, 8$). The other $3$ digits are chosen in order from the remaining $9$: $5 \times P(9, 3) = 5 \times 504 = 2520$.

8. (Challenge) Solve $P(n, 2) = 56$.

Solution

$P(n, 2) = \dfrac{n!}{(n - 2)!} = n(n - 1)$. So $n^2 - n - 56 = 0$, which factors as $(n - 8)(n + 7) = 0$. Since $n$ must be positive, $n = 8$.

Check: $8 \times 7 = 56$. ✓

9. (Challenge) Four boys and three girls stand in a row. In how many ways can they line up if boys and girls must alternate?

Solution

With $4$ boys and $3$ girls, the only alternating pattern is B G B G B G B (starting with a girl would need as many girls as boys). The boys fill their $4$ spots in $4!$ ways and the girls fill their $3$ spots in $3!$ ways:

$$4! \times 3! = 24 \times 6 = 144$$