Permutation and Combination Questions for CAT 2026 — Circular Permutation with Solutions
- Jul 22, 2025
- 3 min read
Updated: Apr 12
PERMUTAION AND COMBINATION Question for CAT
Q.1 The number of ways in which 10 children can sit in a merry-go-round relatively to one another will be
a.10!
b.9!
c.10!/2
d.9!/2
Q.2 In how many ways can we arrange 9 different flowers in a circle to form a garland using these flowers?
a.9!
b.8!
c.9!/2
d.8!/2
Q.3 In how many ways can 16 people be seated across a square table with 4 persons sitting on each side of the square?
a.16!
b.8x15!
c.4x15!
D.15!
Q.4 In how many ways can we arrange 10 people around a rectangular table such that there are 3 chairs on the longer side and 2 on the shorter side ?
a.5 x 9!
b.9!
c.3 x 9!
d.None of the above
Q.5 Let there be 12 people who are needed to be arranged around an equilateral triangular table which has 4 seats on each side. How many such arrangements are possible?
a.11!
b.3x11!
c.4x11!
d.None of the above
Q.6 6 Boys and 6 Girls are to be seated around a circular table so that the person of same gender never sits together. Find the total number of arrangements.
a.21600
b.43200
c.63200
d.86400
Q.7 Find the number of ways in which 5 people A, B, C, D, and E can be seated at a round table, such that A and B always sit together.
a.12
b.24
c.48
d.None of the above
Q.8 Find the number of ways in which 5 people A, B, C, D, and E can be seated at a round table, such that C and D never sit together.
a.12
b.24
c.108
d.None of the above
Q.9 In how many ways can 6 couples be seated around a circular table such that each couple is sitting together?
a.11!
b.6! X 26
c.5! X 26
d.5! X 25
Q.10 In how many ways can 6 Indians and 8 Americans sit across a circular table with 14 equi-spaced chairs such that no two Indians are sitting next to each other?
a.7! X 5!
b.8! X 6!
c.7! X 8P6
d.7! X 6!
Answer for Permutation and Combination for CAT
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
B | D | C | A | C | D | A | A | C | C |
Frequently Asked Questions — Circular Permutation for CAT 2026
What is circular permutation and how does it differ from linear permutation?
In linear permutation, n distinct objects can be arranged in n! ways. In circular permutation, we fix one object's position (since rotations are identical), giving (n-1)! arrangements. For example, 5 people around a round table = (5-1)! = 24 ways.
When do I use (n-1)! vs n!/n for circular permutations?
Both give the same result: (n-1)! = n!/n. Use (n-1)! directly. The logic is: fix one person's seat, then arrange the remaining (n-1) people in (n-1)! ways.
What is the difference between clockwise and anticlockwise in circular permutations?
If clockwise and anticlockwise arrangements are considered different, total arrangements = (n-1)!. If they are considered the same (flippable necklace), divide by 2: arrangements = (n-1)!/2.
How do I solve circular permutation problems with constraints?
For 'A and B must sit together': treat {A,B} as a single unit. Circular arrangements of (n-1) units = (n-2)! x 2! (internal arrangement of A and B). For 'A and B must not sit together': Total circular - (A,B together) = (n-1)! - (n-2)! x 2.
How many ways can 5 men and 5 women sit alternately around a round table?
Fix one man's seat. The remaining 4 men fill 4 seats in 4! ways. The 5 women fill the 5 alternate seats in 5! ways. Total = 4! x 5! = 24 x 120 = 2880.
What is the formula for circular arrangements of beads in a necklace?
For n distinct beads: (n-1)!/2 since flipping the necklace gives the same arrangement. For n beads with some identical, apply the division formula then consider flipping.
How is circular permutation tested in CAT?
CAT tests circular permutation in seating arrangement questions: people around a round table with constraints, beads in a necklace, or circular arrangements of objects. Typically 1 question per year, usually medium difficulty.
What are common mistakes in circular permutation problems?
Common mistakes: (1) Using n! instead of (n-1)! for circular arrangements. (2) Forgetting to divide by 2 for necklace/bracelet problems. (3) Misapplying the 'together' constraint in circular settings. (4) Not fixing a reference point before applying the formula.
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