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# Circular Permutations Formulas

**Formulas of Circular Permutations in Aptitude**

In this Page Circular Permutations Formulas are given for Solving Circular Permutation based questions. Go through this page get clear understanding of the concept of Circular Permutation with the help of Sample Circular Permutation based Questions and Answers.

### Formula

Number of circular-permutations of ‘n’ different things taken ‘r’ at a time:-

**Case 1: **

If clock-wise and anti-clockwise orders are taken as different, then total number of circular-permutations = \frac{^nP_r}{r}

**Case 2: **

If clock-wise and anti-clockwise orders are taken as not different, then total number of circular – permutation = \frac{^nP_r}{2r}

### Formulas & Definition for Circular Permutations:

- The arrangements we have considered so far are linear. There are also arrangements in closed loops, called circular arrangements.
**There are two cases of circular-permutations:** - If clockwise and anti-clock-wise orders are different, then total number of circular-permutations is given by (n-1)!
- If clock-wise and anti-clock-wise orders are taken as not different, then total number of circular-permutations is given by –

\frac{(n-1)!}{2!} - The number of ways to arrange distinct objects along a fixed (i.e., cannot be picked up out of the plane and turned over) circle is (n-1)!

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### Sample Questions of Circular Permutations:

**Question 1:**

A necklace is made using 6 different colored beads. How many distinct necklaces can be formed by rearranging the beads?

**Solution :**

For a necklace, the arrangement is considered the same up to rotation.

Total number of ways to arrange 6 different beads in a line is 6! (factorial), but we need to divide by 6 to account for the circular arrangement of the necklace.

Hence, the total number of distinct necklaces = (6-1)! = 5! = 120

**Question 2:**

In how many ways can 4 people and 3 dogs sit around a circular table with alternating seats for humans and dogs?

**Solution:**

Since the seats are alternating, we need to treat humans and dogs as distinct entities.

First, we will arrange the 4 people in 4! ways around the table. Then, we will arrange the 3 dogs in 3! ways.

However, we must divide by 2 because the dogs are indistinguishable from each other.

Finally, we get the total number of arrangements = \frac{4!\times3!}{2}=36 ways.

**Question 3:**

A team of 8 basketball players is standing in a circle for a team photo. In how many ways can the players arrange themselves if the captain and vice-captain must stand next to each other?

**Solution:**

Consider the captain and vice-captain as a single entity.

Now, we have 7 entities (the pair of captain and vice-captain and the other 6 players) to arrange around a circular table.

Number of ways to arrange 7 entities in a circle is (7-1)! = 6!

However, within the captain-vice-captain entity, there are 2 ways to arrange them. Hence, the total number of arrangements = 2 * 6! = 144 ways

**Question 4:**

In how many ways can 5 books and 4 pens be placed on a circular shelf such that no two pens are adjacent to each other?

**Solution:**

Let’s assume each pair of book and pen together as a single entity.

Now, we have 5 entities (5 pairs) to arrange around the circular shelf.

Number of ways to arrange 5 entities in a circle is (5-1)! = 4!

However, within each pair, there are 2 ways to arrange the book and pen.

###### Hence, the total number of arrangements = 2^5\times4!=2^5\times24=768 ways

**Question 5:**

Five friends – Alice, Bob, Carol, David, and Eve – are going to sit around a circular table. Alice and Bob want to sit together, but David and Eve cannot sit together. In how many ways can they be seated?

**Solution:**

Let’s treat Alice and Bob as a single entity (AB) and David and Eve as a single entity (DE)

Now, we have 4 entities (AB, Carol, DE) to arrange around the circular table. The number of ways to arrange 4 entities in a circle is (4-1)! = 3!

However, within the AB entity, there are 2 ways to arrange Alice and Bob, and within the DE entity, there are 2 ways to arrange David and Eve.

Hence, the total number of arrangements is 2^2 * 3! = 2^2 * 6 = 24 ways

###### 2^2\times3!=2^2\times6=24 ways

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