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PREPINSTA PRIME
How To Solve Cicular Permutation Questions Quickly
How to solve Circular Permutation Questions Quickly
Here on this page you will get the all the helpful Insights on How to Solve Circular Permutation Questions Quickly along with sample Circular Permutation Questions given on this page.
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
- 120
- 12
- 24
- 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
- 2520
- 2400
- 5040
- 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
- 414
- 120
- 240
- 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! x 4!
= 3 x 2 x 1 x 4 x 3 x 2 x 1
= 6 x 24
= 144
Correct option: 4
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Type 2: When clockwise and anticlockwise
Question 1. How many different garlands can be made using 10 flowers of different colors?
Options
- 181440
- 362880
- 145690
- 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
- \frac{10!}{19^2}
- \frac{{19!}^2}{10!}
\frac{19!}{19^2}
\frac{10!}{10^2}
- \frac{10!}{19^2}
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!}{10!}
Now the beads can be arranged in the Circular Fashion in (20-1) = 19 ! ways
Required number of ways = \frac{19!}{10!} \times 19!
= \frac{{19!}^2}{10!}
Correct Option : 2
Question 3. In how many ways can 7 different colors beads be threaded in a string?
Options
- 3600
- 450
- 360
- 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
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