Arranging Letters and Objects: Permutations With Repeated Items (the MISSISSIPPI Problem)
October 21, 2026
If all n items are different, they can be arranged in n! (n factorial) ways. When some items are identical, many of those arrangements look the same, so you divide by the number of ways the identical items can be rearranged among themselves. The formula for n items with repeated groups of sizes k₁, k₂, ... is n! ÷ (k₁! × k₂! × ...). This is called a multinomial coefficient.
The classic example is the word MISSISSIPPI, which has 11 letters: M once, I four times, S four times, and P twice. The number of distinct arrangements is 11! ÷ (1! × 4! × 4! × 2!) = 39,916,800 ÷ (24 × 24 × 2) = 34,650.
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Start by treating every letter as unique, like I₁, I₂, I₃, I₄. There are 11! = 39,916,800 arrangements. But swapping the four I's among themselves does not create a new word: those 4! = 24 swaps all give the same result. The same is true for the four S's (24 swaps) and the two P's (2 swaps). Each distinct word is therefore counted 24 × 24 × 2 = 1,152 times, so we divide.
The formula is easy to remember: the total factorial, divided by the factorial of each repeated group.
More Examples
BANANA has 6 letters with A three times, N twice, and B once: 6! ÷ (3! × 2! × 1!) = 720 ÷ 12 = 60. The word LETTER has 6 letters with E twice and T twice: 6! ÷ (2! × 2!) = 180. A string of beads with 3 red, 2 blue, and 2 green arranged in a line: 7! ÷ (3! × 2! × 2!) = 5,040 ÷ 24 = 210.
A route problem: moving 5 steps east and 3 steps north on a grid takes 8 steps. The number of distinct paths is 8! ÷ (5! × 3!) = 56, which is also C(8, 3).
| Item | Letters or objects | Distinct arrangements |
|---|---|---|
| BANANA | A x3, N x2, B x1 | 60 |
| LETTER | E x2, T x2, L, R | 180 |
| MISSISSIPPI | I x4, S x4, P x2, M x1 | 34,650 |
| 3 red, 2 blue, 2 green beads | 7 objects | 210 |
| Paths 5 east, 3 north | 8 steps | 56 |
Adding Conditions
Often a question adds a constraint. How many arrangements of BANANA start with B? Fix B first, then arrange the remaining AAANN: 5! ÷ (3! × 2!) = 10. How many have all the A's together? Treat AAA as a single block: the items are AAA, N, N, B, giving 4! ÷ 2! = 12 arrangements.
For 'no two identical letters adjacent' questions, the counting is more complex and uses the inclusion-exclusion principle.
Circular Arrangements
Arrangements around a circle, where rotations are considered the same, use (n − 1)! for distinct items. Seating 6 distinct people around a round table can be done in 5! = 120 ways. If reflections are also considered the same (as for a necklace of distinct beads), divide by 2 more: 60.
Where It Applies
Counting distinct codes, license plate patterns, DNA sequences, seating plans, and tournament schedules uses these ideas. In probability, the number of favorable arrangements divided by the total tells you the chance. Check your work by counting a small case by hand to make sure the method counts what you intend.
Frequently Asked Questions
How many ways can the letters of MISSISSIPPI be arranged?
34,650.
What is the formula for permutations with repetition?
n! divided by the product of the factorials of each repeated item count.
How many arrangements does BANANA have?
60.
How do circular arrangements differ?
Rotations count as the same arrangement, so n distinct items have (n − 1)! arrangements.
Is this the same as combinations?
It's related: choosing positions for one type of item is a combination, and the multinomial coefficient is a product of combinations.