Repeated Letters in Word Arrangements: Stop Overcounting

You arrange a set of letter tiles, swap two identical tiles, and look again. Has the word changed?

That tiny question explains why permutations with repeated letters need a correction. A factorial can count the movements of individual tiles even when several movements leave exactly the same visible arrangement.

Imagine two identical coral beads on a thread. You might exchange them a hundred times, but someone looking only at the finished pattern cannot tell. Letter arrangements work the same way: we count the result we can read.

This lesson starts with that idea and builds towards questions involving fixed letters, vowels together, separated repeated letters, and words shorter than the original. Keep a pencil beside you. The useful habit is to record which letters remain before reaching for a formula.

The repeated-letter formula, with its conditions

If you arrange all n letters in a line, and identical letters occur p, q, r, … times, the number of distinct arrangements is:

                 n!
Arrangements = -----------
               p! q! r! ...

For instance, if one letter occurs three times and another occurs twice, divide by 3! x 2!. A letter that appears once contributes 1! = 1, so you can leave that factor unwritten.

This is the standard rule for arranging objects that are not all distinct, explained in NCERT’s Permutations and Combinations chapter, section 6.3.4.

Use the formula with these assumptions clear: every supplied letter is used, arrangements are read from left to right, and copies of the same letter are indistinguishable. A question imposing extra conditions needs an additional step.

A quick reminder: 4! = 4 x 3 x 2 x 1 = 24. Also, 0! = 1 and 1! = 1.

Why do we divide instead of subtract?

Take the four letters of SUNS. There are two S tiles, one U, and one N.

Temporarily mark the S tiles as S1 and S2. Now all four tiles are distinguishable, so there are 4! = 24 labelled arrangements.

Consider just these two:

S1 U N S2
S2 U N S1

Erase the little labels. Both become SUNS.

Every visible arrangement has exactly two labelled versions, because the two S tiles can exchange places in 2! ways. We therefore group the 24 labelled arrangements into pairs:

Distinct arrangements = 24 / 2! = 12

We do not subtract two from 24. The overcount happens inside every group, so subtraction would remove only two entries from a list full of duplicates.

With three identical tiles, each visible arrangement has 3! = 6 labelled versions. With four identical tiles, it has 4! = 24. The divisor counts the invisible rearrangements of identical copies.

Worked example 1: COMMERCE has three repeated pairs

How many distinct arrangements can you make using all the letters of COMMERCE?

Write the word slowly: C, O, M, M, E, R, C, E.

Letter	Number available
C	2
M	2
E	2
O	1
R	1
Total	8

If we labelled every tile, we would count 8! arrangements. Each visible word appears 2! x 2! x 2! = 8 times in that list: the C copies, M copies, and E copies can each exchange places independently.

Arrangements = 8! / (2! x 2! x 2!)
             = 40,320 / 8
             = 5,040

Notice why we do not divide by 6!. The six tiles belonging to repeated groups are not all the same letter. Exchanging a C and an E can change the word. Only exchanges within C, within M, and within E are invisible.

We also do not add the factorials. The two C orders can occur with either M order and either E order, giving a product of choices.

The same factorial correction is presented in OpenStax’s Counting Principles section, under permutations of non-distinct objects.

Check the answer by choosing positions

There is another way to reach 5,040 without starting with an overcounted list.

Choose two of the eight positions for C. Then choose two of the six remaining positions for M, and two of the four remaining positions for E. Finally, put O and R into the last two positions in either order.

Here nCr means choosing r positions from n, without ordering the chosen positions.

C positions: 8C2 = 28
M positions: 6C2 = 15
E positions: 4C2 = 6
O and R:     2! = 2

Total = 28 x 15 x 6 x 2 = 5,040

Once the two C positions have been selected, filling them with C is just one visible choice. We do not multiply by 2! again.

Choosing positions and correcting labelled arrangements count the same final words. The agreement gives you a useful way to check your reasoning.

Worked example 2: BANANAS and the missing factorial trap

The letters of BANANAS are B, A, N, A, N, A, S. There are seven letters: three As, two Ns, one B, and one S.

Arrangements = 7! / (3! x 2!)
             = 5,040 / (6 x 2)
             = 420

A tempting mistake is 7! / (3 x 2) = 840. That divides by the number of copies instead of the number of ways those copies can be rearranged.

The mistake can hide in questions with repeated pairs because 2! happens to equal 2. Three repeated letters expose it: 3! is 6, not 3.

For a fast sense check, imagine a word made of seven As. There is only one visible arrangement, AAAAAAA. The formula gives 7! / 7! = 1, as it should. Dividing by 7 would give 720, which clearly counts things we cannot distinguish.

Fixed positions: use up the fixed letters first

A condition such as “starts with A” changes both the number of empty places and the stock of letters left to fill them.

BANANAS must start with A

Fix an A in the first position:

A _ _ _ _ _ _

The remaining six letters are A, A, N, N, B, S. We now have two As and two Ns to correct for.

Arrangements = 6! / (2! x 2!)
             = 720 / 4
             = 180

There is no extra factor of 3 for choosing which A goes first. The As have no visible labels, so all three choices would create the same set of final strings.

There is also no remaining 3! for the As. One A has already been used; the freely arranged group contains only two.

BANANAS must begin and end with A

Now fix two As:

A _ _ _ _ _ A

The middle contains A, N, N, B, S. Only N repeats there.

Arrangements = 5! / 2!
             = 120 / 2
             = 60

Both results are smaller than the unrestricted total of 420. They also fit each other: beginning and ending with A is more restrictive than beginning with A alone, so its count cannot exceed 180.

A small check using first-letter cases

Every arrangement of BANANAS begins with A, N, B, or S. These cases do not overlap.

First letter	Remaining-letter calculation	Count
A	6! / (2! x 2!)	180
N	6! / 3!	120
B	6! / (3! x 2!)	60
S	6! / (3! x 2!)	60

Adding gives 180 + 120 + 60 + 60 = 420, restoring the unrestricted count. This checks both the fixed-position work and the original total.

Vowels together: count outside and inside the block

How many arrangements of COMMERCE have all the vowels together?

The vowels are O, E, E. Treat them as one block, which we will call V for this calculation. V is a temporary block name, not an extra letter in the answer.

The six objects outside are:

V, C, C, M, M, R

Step 1: arrange the six outer objects

The Cs and Ms still repeat:

Outer arrangements = 6! / (2! x 2!) = 180

Step 2: arrange the vowels inside V

The internal vowel patterns are OEE, EOE, and EEO:

Inner arrangements = 3! / 2! = 3

Step 3: multiply the two counts

Each outer arrangement allows all three internal vowel patterns.

Required arrangements = 180 x 3 = 540

The denominator is doing different work at each stage. Outside, it corrects repeated Cs and Ms. Inside, it corrects the repeated Es. We do not divide by the Es again after combining the stages.

For comparison, putting all three As of BANANAS together gives the outer objects AAA, N, N, B, S. Their count is 5! / 2! = 60. Inside AAA there is only one visible arrangement, since 3! / 3! = 1.

A block has internal arrangements only when its contents can produce different visible orders.

No adjacent copies: place identical letters in gaps

Now arrange BANANAS so that no two As are next to each other. The Ns are allowed to touch; the condition is specifically about A.

First arrange the other four letters, B, N, N, S:

Arrangements of other letters = 4! / 2! = 12

Each arrangement makes five gaps, including the two ends. For the order B, N, N, S, the gaps look like this:

_ B _ N _ N _ S _

To keep the three As apart, choose three different gaps and put one A in each.

Gap selections = 5C3 = 10
Total = 12 x 10 = 120

We do not arrange the three As among the chosen gaps in 3! ways. Once the gaps are chosen, they each receive the same letter. Multiplying by 3! would count every finished arrangement six times.

Each valid string has a unique underlying B, N, N, S arrangement and a unique set of occupied gaps. That is why the multiplication counts every valid result exactly once.

”Not all together” is different from “no two together”

With three As, a string can contain one adjacent pair and a third A elsewhere. For example, AABNANS has all the correct letters, contains AA, and does not contain AAA.

So it belongs to the “As not all together” group, but fails the “no two As adjacent” condition.

Condition on BANANAS	Calculation	Count
All three As together	5! / 2!	60
As not all together	420 - 60	360
No two As adjacent	12 x 5C3	120
At least one adjacent A pair	420 - 120	300

Subtracting the AAA-block count does not remove strings such as AABNANS. Read the wording before choosing a complement.

For this example, there are enough gaps to separate the three As. More generally, m other letters make m + 1 gaps. If the number of identical copies you want to separate exceeds that, separation is impossible.

Shorter words: the full-word denominator is not enough

Suppose the question asks for three-letter strings from LEVEL, using each supplied tile at most once.

LEVEL contains two Ls, two Es, and one V. Because we are not using all five tiles, the selected letters can have different repetition patterns.

A formula such as 5P3 / (2! x 2!) cannot handle those changing patterns. It would give 15, but the correct answer is 18.

The distinction between arranging an entire collection with repeated items and taking only some of its items is formalised in Martin Bays’s notes on multisets. For a small word, an organised case list makes the distinction easy to use.

Case 1: three different letters

Use L, E, and V. There is one selection of letter types and 3! = 6 orders:

LEV, LVE, ELV, EVL, VLE, VEL

Do not multiply for choosing one of the two L tiles or one of the two E tiles. Their copies are identical in the word we read.

Case 2: a repeated pair and one different letter

Only L and E can provide a pair. The possible collections and their arrangements are:

Selected letters	Distinct strings	Count
L, L, E	LLE, LEL, ELL	3
L, L, V	LLV, LVL, VLL	3
E, E, L	EEL, ELE, LEE	3
E, E, V	EEV, EVE, VEE	3

Each collection gives 3! / 2! = 3 orders, so this case contributes 12.

Three identical letters are impossible: neither L nor E is available three times, and V is available only once.

Total three-letter strings = 6 + 12 = 18

Every valid string either has three different letters or has one repeated pair. The cases cover everything and do not overlap.

Four-letter strings from LEVEL

This time, using four letters means leaving out one tile. Organise cases by the letter type omitted:

Omitted type	Letters used	Count
L	L, E, E, V	4! / 2! = 12
E	L, L, E, V	4! / 2! = 12
V	L, L, E, E	4! / (2! x 2!) = 6

The total is 12 + 12 + 6 = 30.

Leaving out the first L or the second L is the same visible selection. That is why the first row appears once. These three cases also have different letter counts, so no finished string can appear in two rows.

The general habit is simple: list the possible selected letter counts, arrange each selection with its own denominator, and add the non-overlapping counts.

Fixed repeated letters versus unlimited reuse

The phrase “repetition allowed” can describe a different problem from rearranging the letters of a word.

Compare these two instructions:

Make three-letter strings from the tiles in LEVEL, without replacement. You have L twice, E twice, and V once. We found 18 strings.

Make three-letter strings from the alphabet L, E, V, with unlimited reuse. Each position has three choices, independently of the others:

3 x 3 x 3 = 3^3 = 27

Now LLL, EEE, and VVV are allowed. So are strings with two Vs. Those were impossible with the original five tiles.

The number 5 in LEVEL counts tiles. The number 3 counts different symbols. Decide which supply the question gives you before using a power or a factorial.

If the starting question is whether order matters at all, work through the five-question permutation-or-combination test first.

A reliable answer-writing routine

Use these steps when you practise:

1. State what counts as different. For ordinary letter strings, position matters and identical copies have no separate identities.
2. Record the supply. Write each letter and its frequency, then check the total.
3. Apply the condition. Fix letters, create a block, arrange separators, or split shorter selections into cases.
4. Count the task that remains. Use the current number of positions and the current repetition counts.
5. Explain each multiplier or divisor. It should correspond to a choice or an overcount you can describe.
6. Check the scale. A restricted count cannot exceed the total for the same word and length.

A fractional answer to a counting question signals a mistake, but an integer answer does not prove correctness. The incorrect LEVEL calculation gave the perfectly ordinary integer 15. Checking the cases is stronger than checking whether the answer looks tidy.

Practice questions

Try these before looking at the answers. Unless stated otherwise, use all supplied letters in a straight line, treat identical letters as indistinguishable, and allow strings without a dictionary meaning.

1. How many distinct arrangements can you make from COCOA?
2. How many arrangements of COCOA start with C?
3. How many arrangements of ASSESS are possible?
4. How many arrangements of COMMERCE have the two Ms together?
5. How many arrangements of COMMERCE keep the two Cs apart?
6. How many arrangements of BANANAS put all three As in odd-numbered positions?
7. How many two-letter strings can you make from LEVEL without replacement?
8. A display uses two identical red cards, three identical blue cards, and one gold card. How many colour sequences are possible in a row?

Answers with working

1. COCOA: 30. There are five letters, with C twice and O twice. The count is 5! / (2! x 2!) = 30.

2. COCOA starting with C: 12. Fix C first. The remaining letters are C, O, O, A, giving 4! / 2! = 12. There is no factor of 2 for choosing the first C.

3. ASSESS: 30. Count carefully: S appears four times. With six letters altogether, the answer is 6! / 4! = 30.

4. COMMERCE with MM together: 1,260. Treat MM as one object. Arrange it with C, C, E, E, O, R, giving 7! / (2! x 2!) = 1,260. The MM block has one internal arrangement.

5. COMMERCE with Cs apart: 3,780. The two Cs together give a CC block and six other letters, with M and E still repeated. That count is 7! / (2! x 2!) = 1,260. Subtract from the total: 5,040 - 1,260 = 3,780. With exactly two Cs, together and apart exhaust the possibilities.

6. BANANAS with As in odd positions: 48. The odd positions are 1, 3, 5, and 7. Choose three for A in 4C3 = 4 ways. Fill the remaining four positions with B, N, N, S in 4! / 2! = 12 ways. Multiply to get 48. Odd positions do not all have to contain A; the instruction only restricts where the three As may go.

7. Two-letter strings from LEVEL: 8. Two different symbols give 3P2 = 6: LE, LV, EL, EV, VL, and VE. Add LL and EE. VV is unavailable because there is only one V tile.

8. Six-card colour display: 60. The answer is 6! / (2! x 3!) = 60. Treat the gold card as a single distinct item. This is the repeated-letter idea with colours replacing symbols.

Sources and further reading

The explanations and practice variations above use the standard counting principles in these references:

• NCERT, Mathematics: Permutations and Combinations, reprint 2026-27, particularly section 6.3 on permutations and repeated objects.
• Jay Abramson, OpenStax: Counting Principles, from Algebra and Trigonometry, published 2015, for multiplication, combinations, and non-distinct arrangements.
• Martin Bays: Multisets, undated lecture notes, for the distinction between an entire repeated collection and a shorter ordered selection.

Frequently asked questions

What is the formula for permutations with repeated letters?

When all n letters are arranged without additional restrictions, divide n! by the factorial of each repeated letter’s frequency. If the counts are p, q, and r, use n! / (p! x q! x r!).

Why do we divide by the factorial of each repetition count?

A factorial initially treats the copies as distinguishable. If a letter occurs p times, its copies can exchange places in p! ways without changing the visible word. Division removes that repeated counting.

Do I divide by 3 or 3! when a letter occurs three times?

Divide by 3! = 6. You are correcting for six orders of three temporarily labelled copies, not simply for the presence of three tiles.

How many arrangements does COMMERCE have?

There are 5,040. COMMERCE has eight letters, with C, M, and E each occurring twice, so the calculation is 8! / (2! x 2! x 2!).

Do repeated letters have to be together in the original word?

No. Their original positions do not affect the unrestricted total. Both adjacent copies and separated copies belong to the same repeated-letter count.

Do arrangements have to be meaningful English words?

Usually, a word-arrangement question counts all permitted strings, whether or not they have meanings. If the question specifically requires meaningful words, the factorial formula alone cannot identify them.

What changes when the first letter is fixed?

Use up that letter and recount the remaining stock. BANANAS starting with A leaves six letters, including two As and two Ns, so the answer is 6! / (2! x 2!) = 180.

Should I multiply by the number of identical letters available for a fixed position?

No. Choosing among identical copies does not produce different visible results. An extra multiplier is justified only when the alternatives are distinguishable in the outcome being counted.

How do I handle repeated vowels that must stay together?

Treat the vowels as one outer object, count the outer arrangements, and multiply by the distinct orders inside the vowel block. Correct for repeated letters separately at the stage where they are arranged.

Is “not all together” the same as “no two adjacent”?

No, when there are three or more copies. Two copies can touch while another sits elsewhere. For the three As in BANANAS, “not all together” gives 360 arrangements, while “no two adjacent” gives 120.

Can I use the full-word denominator for a shorter word?

Not automatically. Shorter selections can have different repetition counts. For three-letter strings from LEVEL, count the all-different selection and each repeated-pair selection separately, giving 18 altogether.

Are different letter strings always equally likely when tiles are drawn randomly?

Not always. With all tiles used in a uniformly random ordering, every distinct full-length string has the same number of labelled versions and is equally likely. For shorter draws, strings with different letter counts can have different probabilities. From LEVEL, for example, the probability of drawing LL in order is (2/5) x (1/4) = 1/10, while that of LE is (2/5) x (2/4) = 1/5. Count distinct strings and calculate their probabilities as separate tasks.