Learn: arrange four letter tiles
You have four tiles: A, A, B, B. Place every tile in a row. Swapping the two A tiles does not produce a new visible word. How many different rows can you make?
The six rows are AABB, ABAB, ABBA, BAAB, BABA, BBAA. Order matters: AABB and ABAB are different. But you must use exactly two A’s and two B’s.
Vocabulary before the formula
- An arrangement puts objects into ordered positions.
- Identical items look the same for this counting question; exchanging them does not create a distinguishable arrangement.
- A factorial, written n!, multiplies the positive whole numbers from n down to 1. For example, 4! = 4 × 3 × 2 × 1 = 24. By convention, 0! = 1.
Why divide when items are identical?
If you temporarily label the tiles A₁, A₂, B₁, B₂, there are 4! = 24 orders. Every visible row appears 2! × 2! = 4 times among those labelled orders: the A labels can swap, and the B labels can swap. Removing that overcount gives 24 ÷ 4 = 6 visible rows.
For n total items, with group sizes n₁, n₂, … adding up to n, the count is n! ÷ (n₁! × n₂! × …). Each denominator group contains items identical to one another. The exclamation mark means factorial, not excitement or a unit.
Worked example: repeated colours
Arrange eight balls in a row: three red, three blue and two green. Balls of the same colour are indistinguishable.
- Count all the balls: n = 3 + 3 + 2 = 8.
- Account for repeated colours: 8! ÷ (3! × 3! × 2!).
- Calculate: 40,320 ÷ (6 × 6 × 2) = 40,320 ÷ 72 = 560.
- Interpret: there are 560 distinct colour rows using exactly those eight balls.
Explore: fixed tiles or reusable choices?
Now consider a four-character code made only from A and B, with either letter allowed in every position. AAAA is now allowed. That is a different question from arranging AABB.
4! ÷ (2! × 2!) = 6. Every row uses exactly two A’s and two B’s.
AABB · ABAB · ABBA · BAAB · BABA · BBAA
For reusable choices, multiply the choices at each position: 2 × 2 × 2 × 2 = 2⁴ = 16. More generally, k choices available at each of r positions give kʳ sequences, provided no extra restriction removes choices. A six-character code using A, B or C has 3⁶ = 729 possibilities under those assumptions.
Worked example: MISSISSIPPI
The word contains 11 letters: M once, I four times, S four times and P twice. The number of distinguishable rearrangements is 11! ÷ (1! × 4! × 4! × 2!) = 39,916,800 ÷ 1,152 = 34,650. These are letter sequences; they do not all need to be dictionary words.
Practice: state the restriction first
1. How many rearrangements of BALLOON are possible?
There are seven letters with two L’s and two O’s. 7! ÷ (2! × 2!) = 5,040 ÷ 4 = 1,260.
2. How many four-digit codes use digits 0–9 if repeats and a leading zero are allowed?
Each of four positions has ten choices. 10⁴ = 10,000. The leading-zero condition matters: a code is not necessarily a four-digit number.
3. How many rows use exactly A, A, A, B?
4! ÷ 3! = 4: AAAB, AABA, ABAA and BAAA. Equivalently, choose one of four positions for B.
4. Five books include two identical novels, two identical dictionaries and one different book. Count the shelf orders.
5! ÷ (2! × 2!) = 120 ÷ 4 = 30. This assumes copies described as identical are indistinguishable for the task.
Review: choose the model before calculating
Ask whether order matters, whether you must use a fixed collection, and whether items can be reused. Fixed identical items lead to factorial division; reusable choices in each slot lead to multiplication. A few correct answers are a check of understanding, not proof of mastery.
Optional reference: OpenStax counting principles. Continue with Algebra 2 or AP Statistics.