Fundamentals of Business Mathematics and Statistics · Permutation and Combinations

Permutations with Repetition and Restrictions Explained

Updated 10 October 2026 · Fact-checked

These are arrangement problems with a twist. With identical objects, divide n! by the factorial of each repeat count. For circular arrangements of n distinct objects, use (n−1)!. For "together" cases, glue the items into one block and multiply by the block's internal arrangements. For "never together", subtract the together cases from the total.

Understand Permutations with Repetition and Restrictions

A permutation is an arrangement where order matters. For n distinct objects in a row, the number of arrangements is n!. The rules in this topic change that basic count when something special is true about the objects or the positions.

First, identical objects. In the word LEVEL, the two L's look the same and so do the two E's. Swapping the two L's gives the same word. If you use n!, you count each real arrangement more than once. So you divide by the number of ways the identical objects can be swapped among themselves: 2! for the L's and 2! for the E's.

Second, repetition allowed. Here you may reuse an item, such as digits in a PIN. Each position is filled independently, so the count is a product of choices, not a factorial. With n choices for each of r positions, you get n^r.

Third, circular arrangements. Around a round table there is no fixed start or end. Rotating everyone by one seat gives the same arrangement. So you fix one object to remove the rotations and arrange the rest in (n−1)! ways. A linear arrangement has a first and last place, so rotations give different results and the count is n!.

Fourth, restrictions. For items that must be together, treat them as one block, arrange the block with the others, then arrange inside the block. For items that must not be together, find the total arrangements and subtract the together arrangements.

Key formulas to remember

Arrangements with identical objects: n! ÷ (p! × q! × r! × …); n objects in total, of which p are of one kind, q of another, r of another. Objects that occur once need no division.
Repetition allowed: n^r; Number of ways to fill r positions when each position can be filled in n ways and reuse is allowed.
Circular permutation of distinct objects: (n − 1)!; Applies when only the relative positions matter and clockwise and anticlockwise are counted as different.
Circular permutation when clockwise and anticlockwise are the same: (n − 1)! ÷ 2; Used for necklaces, garlands and similar cases that can be flipped over. Valid for n ≥ 3.
Items always together (linear): (n − k + 1)! × k!; k specified distinct items form one block among n distinct objects.
Items never together: Total arrangements − Arrangements with the items together; Works for both linear and circular cases. Use the matching total and together counts.
Gap method for never together (linear): (n − k)! × (n − k + 1)P k; Arrange the other n − k objects first, then place the k items in separate gaps. There are n − k + 1 gaps.

How to solve Permutations with Repetition and Restrictions questions

Use the same short routine for every question in this topic. It stops you from picking a formula too early.

1Read the question and decide whether the arrangement is in a line or in a circle.
2List the objects. Check whether any are identical and note how many of each kind.
3Check whether repetition of an object is allowed, such as digits reused in a number.
4Mark any restriction: must be together, must not be together, or a fixed position.
5For a restriction, build the count: block method for together, gap method or subtraction for never together.
6Apply the formula: n!, n!/(p!q!…), n^r or (n − 1)! as needed, and multiply the stages of the task.
7Simplify with cancelling before multiplying, and check the answer is a whole number that matches one option.

Quickest way: Block, gap and cancel

When to use it: Use this for MCQs on words, seating and number-forming questions where you need the count fast and the numbers are small.

For identical letters, write n! over the repeat factorials and cancel by hand before multiplying.
For together, shrink the group to one object. Count the new total, take its factorial, and multiply by the group's own factorial.
For never together, compute total minus together. This is usually quicker than the gap method.
For circular seating, fix one person first and use (n − 1)!. For a block in a circle, count the block as one object, then use (m − 1)! with m objects, and multiply by the block's internal arrangements.
Check the last digit or size of your result against the options to remove wrong choices quickly.

Common mistakes in Permutations with Repetition and Restrictions

Using n! for a word with repeated letters without dividing.
Students treat identical letters as different.
Fix: Count each letter first. Divide n! by the factorial of each repeat count.
Forgetting the internal arrangements of a block in a together question.
After gluing items into one block, students stop at the block count.
Fix: Always multiply by k! for the k items inside the block, if they are distinct.
Using n! for circular seating.
Students carry the linear rule over to the table.
Fix: In a circle, rotations are the same. Fix one object and use (n − 1)!.
Subtracting the wrong quantity in a never-together question.
Students mix linear totals with circular together counts, or the reverse.
Fix: Use the same type of arrangement for both: linear total with linear together, circular total with circular together.
Using a factorial when repetition is allowed.
Students assume every arrangement question uses n!.
Fix: If an item can be reused, each place has the same number of choices. Use n^r.
Dividing by 2 in every circular question.
Students memorise (n − 1)!/2 and apply it blindly.
Fix: Divide by 2 only when the question says clockwise and anticlockwise are the same, as with garlands or necklaces.

Worked examples

Example 1

How many different arrangements can be made from all the letters of the word BANANA, and in how many of them do the three A's stay together?

Show the solution

BANANA has 6 letters: B once, A three times, N twice.
Total arrangements = 6! ÷ (3! × 2!) = 720 ÷ (6 × 2) = 720 ÷ 12 = 60.
For the A's together, treat AAA as one block. The objects are now: block, B, N, N, which is 4 objects with N repeated twice.
Arrangements = 4! ÷ 2! = 24 ÷ 2 = 12.
The A's are identical, so the block has only one internal arrangement.

Answer: Total arrangements = 60; with the three A's together = 12.

Example 2

Five boys and three girls sit in a row. In how many ways can they sit if the three girls must not sit together? Also, in how many ways can 6 people sit around a round table?

Show the solution

Total arrangements of 8 distinct people in a row = 8! = 40,320.
Girls together: treat the 3 girls as one block. Objects = 5 boys + 1 block = 6, giving 6! = 720 ways.
Inside the block the girls can be arranged in 3! = 6 ways.
Arrangements with girls together = 720 × 6 = 4,320.
Girls not together = 40,320 − 4,320 = 36,000.
For the round table, fix one person and arrange the other 5: (6 − 1)! = 5! = 120.

Answer: Girls not together: 36,000 ways. Six people around a round table: 120 ways.

Exam tips

Scan the question for the words "all letters", "together", "not together", "round table" and "repetition allowed". Each one points to a specific rule.
Cancel factorials by hand before multiplying. This is faster and avoids big-number slips on an OMR paper.
For never together, take the subtraction route unless the question gives gaps clearly. It is easier to check.
Look at the options. Wrong answers often come from forgetting the block's internal arrangements or the circular (n − 1)!, so test your result against those traps.
There is no negative marking, so always mark an option. If stuck, eliminate those that are not whole numbers or are bigger than n!.

Practice questions from Permutation and Combinations

Permutations with Repetition and Restrictions in other exams

The same ground in other exams, if you are preparing for more than one or want another angle on it.

CA FoundationPermutations with Repetition and Identical Objects

Permutations with Repetition and Restrictions: frequently asked questions

What is the formula for permutations when letters are repeated?

Use n! ÷ (p! × q! × r! × …), where n is the total number of letters and p, q, r are the counts of each repeated letter. Letters that appear once need no division.

What is the difference between linear and circular permutation?

In a linear arrangement, positions are fixed in a line, so n distinct objects give n! arrangements. In a circular arrangement, rotations look the same, so you fix one object and get (n − 1)!.

How do I solve questions where items must always be together?

Treat the items as one block and arrange the block with the remaining objects. Then multiply by the number of ways to arrange the items inside the block. For k distinct items among n objects this is (n − k + 1)! × k!.

How do I find arrangements where items are never together?

Find the total number of arrangements, then subtract the number of arrangements in which the items are together. Keep both counts of the same type, linear or circular.