# How To Solve Cicular Permutation Questions Quickly

## How to solve Circular Permutation Questions Quickly

Circular Permutation of an arrangement in a closed loop can be calculated by the formula ,(n-1) ! where n is the number of items. ### Definition & How to Solve Circular Permutation

• Combination is an arrangement of objects where order does not matter.
• There are also arrangements in closed loops, called circular arrangements

### Type 1: Find the greatest or smallest number.

Question 1. Anuradha invited her 5 friends for dinner. In how many ways she can make them sit around a circular table?

Options

1. 120
2. 12
3. 24
4. 72

Solution:    Number of arrangements possible = (n − 1)!

= (5 – 1)!

= 4!

= 24

Correct option: 3

Question 2. A gardener wants to plant some Neem trees around a circular pavement. He has 7 different size of Neem trees. In how many different ways can the Neem tree be planted?

Options

1. 2520
2. 2400
3. 5040
4. 720

Solution:    Number of arrangements possible = (n − 1)!

= (7 – 1)!

= 6!

= 720

Correct option: 4

Question 3. In how many ways can 4 men and 4 women be seated at a circular table so that no 2 women sits together?

Options

1. 414
2. 120
3. 240
4. 144

Solution:     4 men may be seated in 3! ways, leaving one seat empty. Then at remaining 4 seats, 4 women can sit in 4! ways.

= 3! * 4!

= 3 * 2 * 1 * 4 * 3 * 2* 1

= 6 * 24

= 144

Correct option: 4

### Type 2: When clockwise and anticlockwise

Question 1. How many different garlands can be made using 10 flowers of different colors?

Options

1. 181440
2. 362880
3. 145690
4. 5040

Solution:    Number of arrangements possible = $\frac{1}{2}$ × (n-1) !

= $\frac{1}{2}$ × (10-1) !

= $\frac{1}{2}$ × 9 !

= $\frac{1}{2}$ × 362880

= 181440

Correct option: 1

Question 2. How many necklace of 10 beads each can be made from 20 beads of different colors?

Options

1. $\frac{10!}{19^2}$
2. $\frac{{19!}^2)}{10!}$
3. $\frac{19!}{19^2}$
4. $\frac{10!}{10^2}$

Solution:    In case of necklace the clockwise or anticlockwise arrangements are not different. Therefore, the required ways

= $\frac{^{20}P_{10}}{10 ×2}$

= $\frac{20! }{10! × 10 ×2}$

= $\frac{{19!}^2)}{10!}$

Correct Option : 2

Question 3. In how many ways can 7 different colors beads be threaded in a string?

Options

1. 3600
2. 450
3. 360
4. 540

Solution:    As necklace can be turned over, clockwise and anti-clockwise arrangements are the same. Therefore, Number of arrangements possible

= $\frac{1}{2}$ × (n-1)!

= $\frac{1}{2}$ × (7-1)!

= $\frac{1}{2}$ × 6!

= 360

Correct option: 3