Quantitative Aptitude · Permutations and Combinations

Permutations with Repetition and Identical Objects

Updated 1 October 2026 · Fact-checked

Permutations with identical objects count the distinct arrangements when some items are alike. Divide n! by the factorial of each group of alike items: n! ÷ (p! q! r!). When repetition is allowed, each of r places can be filled in n ways, giving n^r. For dictionary rank, count the words that come before.

Understand Permutations with Repetition and Identical Objects

With distinct objects, n items can be arranged in n! ways. Now suppose some items are alike, as in the word BALL. Swapping the two L's gives the same word, so n! counts every real arrangement more than once.

If p items are alike, they can be shuffled among themselves in p! ways, and none of those shuffles changes the word. So you divide by p!. If there are several groups of alike items, divide by the factorial of each group.

A second situation is repetition allowed. Here you fill places one by one, and the same item can be used again. Every place has the same number of choices, so you multiply. Filling r places from n items gives n × n × ... × n = n^r.

A third situation is rank in a dictionary. You list the letters in alphabetical order and count how many words come before the given word. You fix letters from the left. At each position, count the words that start with a smaller letter, then move to the next position. Add 1 at the end.

Keep two ideas apart. Alike objects mean you divide. Repetition allowed means you raise to a power. Reading the question carefully for which one applies is most of the topic.

Key formulas to remember

Arrangements with identical objects: n! ÷ (p! × q! × r! ...); n is the total number of objects. p, q, r are the sizes of the groups of alike objects. Letters that appear once contribute 1! and can be ignored.
Arrangements with repetition allowed: n^r; r places, each filled in n ways. Use it for codes, PINs, and numbers where digits can repeat.
Arrangements of n objects taken r at a time, repetition allowed: n^r; Same rule. It is not nPr, because nPr does not allow reuse.
Alike items kept together: Treat the group as one block, then arrange the blocks with the formula for identical objects; Example: if two E's must be together, count them as one unit, so the unit is not divided by 2! again.
Rank of a word in a dictionary: Rank = (sum of words starting with smaller letters at each position) + 1; At each position, count words with each smaller unused letter placed there. The remaining letters are arranged using n! ÷ (p! q! ...).
Numbers from digits with repetition allowed: Choices for first place (non-zero) × choices for other places; If 0 is among the digits, the first place of a multi-digit number cannot be 0.

How to solve Permutations with Repetition and Identical Objects questions

Use this method for any question on arranging letters, digits or objects when some are alike or repetition is allowed.

1Read the question and decide the type: alike objects (fixed set of items) or repetition allowed (choices stay the same at each place).
2For alike objects, count the total letters n. List each repeated letter and how many times it appears.
3Write the formula n! ÷ (p! q! ...) and cancel factors before multiplying. Do not expand the big factorial.
4If there is a condition (together, first letter fixed, vowels at odd places), apply it first. Fix the letter or form the block, then count the remaining arrangements.
5For repetition allowed, fill the restricted place first (for example, the first digit cannot be 0 or the last digit must be even). Then multiply the choices for the other places.
6For dictionary rank, write the letters in alphabetical order. Go left to right and, at each position, count the words starting with each smaller unused letter. Add these counts and add 1.
7Check that your answer is a whole number and is not more than n! or the maximum possible count, then match it with an option.

Quickest way: Cancel first, then match the options

When to use it: Use this under time pressure in MCQs on letter arrangements, number formation and rank questions.

Count the letters and the repeats in under 10 seconds. Write them as n with repeats like (4, 4, 2).
Cancel the largest repeated factorial against n! first. For example, 11! ÷ 4! leaves 11×10×9×8×7×6×5.
Check the last digit. If the numerator has a factor of 10 or the product is clearly even, that eliminates options ending in odd digits.
Eliminate options larger than n! (for alike objects) or larger than n^r (for repetition). Many wrong options are the n! value itself.
For rank questions, first compute the total number of words. If your rank is bigger than this total, you made an error.
Skip a rank question with 7 or more letters and many repeats unless you have spare time. A wrong answer costs 0.25 marks, and these questions take longest.

Common mistakes in Permutations with Repetition and Identical Objects

Using n! and forgetting to divide by the factorial of alike letters.
Students practise distinct-object problems first and apply n! automatically.
Fix: Before using n!, scan the word for repeated letters. Write the counts of each letter next to the word.
Dividing by the number of repeated letters instead of its factorial, such as dividing by 4 instead of 4!.
The idea of 'dividing for repeats' is remembered, but not the exact rule.
Fix: Alike items in a group of p can be shuffled in p! ways. Divide by p!, never by p.
Using nPr or n! when repetition is allowed.
Students link the word 'arrangement' with permutations without repetition.
Fix: Ask: can the same item be used again? If yes, multiply the number of choices for each place, giving n^r.
Allowing 0 in the first place when forming numbers.
Students count all 10 digits for every place.
Fix: Fill the first place first with only non-zero digits. Then fill the other places.
In a rank question, not using the alphabetical order of the letters or counting repeated letters as separate starting choices.
Students treat identical letters as different when listing smaller letters.
Fix: List distinct letters only in alphabetical order. At each position, count only the smaller distinct letters not already used up, then adjust the remaining letters for the count.
Forgetting to add 1 at the end of a rank question.
Students count the words before the given word and stop.
Fix: Rank = number of words before it + 1. Write '+1' as the last line of your working.

Worked examples

Example 1

How many distinct arrangements can be made from the letters of the word MISSISSIPPI? (a) 34,650 (b) 69,300 (c) 17,325 (d) 3,99,16,800

Show the solution

The word has 11 letters: M once, I four times, S four times, P two times.
Check the total: 1 + 4 + 4 + 2 = 11.
Number of arrangements = 11! ÷ (1! × 4! × 4! × 2!).
11! ÷ 4! = 11 × 10 × 9 × 8 × 7 × 6 × 5. Stepwise: 11 × 10 = 110; 110 × 9 = 990; 990 × 8 = 7,920; 7,920 × 7 = 55,440; 55,440 × 6 = 3,32,640; 3,32,640 × 5 = 16,63,200.
Divide by the rest: 4! × 2! = 24 × 2 = 48.
16,63,200 ÷ 48 = 34,650.

Answer: (a) 34,650

Example 2

The letters of the word LEVEL are arranged in all possible ways and listed in dictionary order. What is the rank of LEVEL? (a) 17 (b) 16 (c) 18 (d) 13

Show the solution

Letters in alphabetical order: E, E, L, L, V. Total words = 5! ÷ (2! × 2!) = 30.
Position 1: the word starts with L. Smaller letter: E. Words starting with E use the remaining E, L, L, V: 4! ÷ 2! = 12 words.
Fix L. Remaining letters: E, E, L, V. Position 2 is E. There is no smaller letter than E, so add 0.
Fix LE. Remaining: E, L, V. Position 3 is V. Smaller letters: E and L.
Starting LEE: remaining L, V give 2! = 2 words. Starting LEL: remaining E, V give 2! = 2 words. Total 4.
Fix LEV. Remaining: E, L. Position 4 is E, the smallest, so add 0. Position 5 is L, fixed.
Words before LEVEL = 12 + 0 + 4 + 0 = 16. Rank = 16 + 1 = 17.

Answer: (a) 17

Example 3

How many 3-digit even numbers can be formed using the digits 0, 1, 2, 3, 4 if digits can be repeated? (a) 60 (b) 100 (c) 75 (d) 48

Show the solution

Fill the restricted places first. Hundreds place cannot be 0, so it has 4 choices: 1, 2, 3, 4.
Units place must be even: 0, 2 or 4. That gives 3 choices.
Tens place has no restriction and repetition is allowed, so it has 5 choices.
Total = 4 × 5 × 3 = 60.

Answer: (a) 60

Exam tips

Look for repeated letters in the very first read of the word. Questions often use words like MISSISSIPPI, BANANA, COMMITTEE, or LEVEL.
Questions with conditions such as 'vowels together' or 'first and last letters fixed' are common. Do the condition first, then divide for the repeats left in the remaining letters.
Wrong options are usually the n! value, the value without one factorial, or a value divided by p instead of p!. Do not pick the first option that looks close.
For digit problems, check if the question says 'repetition allowed' or 'not allowed', and whether numbers must be even, odd or greater than a given value.
Keep rank questions for the end of the paper. They need careful steps and one slip changes the answer.

Practice questions from Permutations and Combinations

Permutations with Repetition and Identical Objects: frequently asked questions

What is the formula for permutations when some objects are identical?

If n objects include p alike of one kind, q alike of another and r alike of a third, the number of distinct arrangements is n! ÷ (p! q! r!). Objects that appear once do not change the answer. Cancel the large factorial before multiplying.

How do I find the number of ways to arrange letters of a word with repeated letters?

Count the total letters and the number of times each letter repeats. Divide the factorial of the total by the factorial of each repeat count. For BANANA, 6! ÷ (3! × 2!) = 60.

What is the difference between permutations with identical objects and permutations with repetition allowed?

With identical objects, you have a fixed set of items and some are alike, so you divide by factorials. With repetition allowed, you can use the same item many times and multiply the choices for each place, giving n^r.

How do I find the rank of a word in a dictionary?

Arrange the distinct letters alphabetically. Move left to right, and at each position count the words that start with a smaller unused letter, using the remaining letters. Add all these counts and add 1.

Do I need to cancel factorials, or can I use a calculator?

CA Foundation objective papers do not allow a calculator, so cancel factorials first. This keeps numbers small and reduces calculation errors.