A group of five people consisting a couple are to be seated on a round table for a meeting. What is the total number of ways in which the seating arrangement can be made so that the couple do NOT sit next to each other?

Q: What is the answer to CAPF 2024 Paper I Q64?

C. Correct — C, 12. Around a circular table only relative positions matter, so n people can be seated in (n − 1)! ways; for five people that is 4! = 24. Now count the arrangements the question wants to exclude. Tie the couple together as a single block, leaving four units to seat in a circle: (4 − 1)! = 3! = 6 ways, and the couple can swap within the block, giving 6 × 2 = 12. Subtracting, 24 − 12 = 12 arrangements keep the couple apart. Around a table of five, exactly half the seatings separate a given pair. (Provisional — pending the official UPSC key.)

(a)24
(b)18
(c)12
(d)6

Answer

Why

Correct — C, 12. Around a circular table only relative positions matter, so n people can be seated in (n − 1)! ways; for five people that is 4! = 24. Now count the arrangements the question wants to exclude. Tie the couple together as a single block, leaving four units to seat in a circle: (4 − 1)! = 3! = 6 ways, and the couple can swap within the block, giving 6 × 2 = 12. Subtracting, 24 − 12 = 12 arrangements keep the couple apart. Around a table of five, exactly half the seatings separate a given pair.

Why the others are wrong

(a)24 — 24 is the total number of circular arrangements of five people with no restriction at all. It is the figure to subtract from, not the answer.
(b)18 — 18 would leave only 6 arrangements with the couple together, which is the block count before allowing the two to swap places.
(d)6 — 6 is (4 − 1)!, the number of ways to seat four units in a circle. It counts the couple as one person and omits the factor of 2 as well as the subtraction.

Concept

A circular arrangement has no fixed first seat, so rotating everyone by one chair produces the same arrangement. Fixing one person removes that freedom and leaves (n − 1)! orders for the rest. Restrictions are handled by the same block-and-subtract method used in a row: glue the constrained pair into one unit, count, multiply by the internal arrangements of the unit, and subtract from the unrestricted total.

Compare the row version to see what the circle changes. Five people in a row give 5! = 120 arrangements, of which 4! × 2 = 48 keep a couple adjacent, leaving 72 apart. In a circle the totals drop by a factor of five and the ratio changes, because a circle of five gives each person two neighbours rather than an average of fewer at the ends. Around this table half of all seatings separate the pair, which is a useful sanity check on the answer.

Key facts

n people can be seated around a circular table in (n − 1)! ways.
Five people give 4! = 24 circular arrangements.
Treating the couple as one block gives 3! = 6 circular orders, doubled to 12 for the internal swap.
Arrangements with the couple apart = 24 − 12 = 12.
In a row the same five people give 120 arrangements, 48 with the couple together and 72 apart.

Total minus the unwanted

All circular seatings→(5 − 1)! = 4! = 24

Couple treated as one block→(4 − 1)! = 6 circular orders

Couple can swap→6 × 2 = 12 seatings with them together

Couple apart→24 − 12 = 12

Counting the forbidden cases and subtracting is faster than counting the allowed ones directly.

Study next

Circular permutations and why one factor is lostArrangements with adjacency required or forbiddenNecklace and garland arrangements, where reflections coincidePermutations of objects in a row with restrictionsThe complement principle in counting

Common traps

Using 5! instead of 4! and treating the circle like a row.
Counting the couple as a block and forgetting the internal factor of 2.
Reporting the arrangements with the couple together rather than apart.

A circular-permutation item with a negative condition, which is the standard cue to count the total and subtract the forbidden arrangements.

Related PYQs

UPSC_1998_GS1_Q1401998Open & attempt →

A, B, C, D, E and F, not necessarily in that order, are sitting on six chairs regularly placed around a round table. It is observed that A is between D and F, C is opposite D, and D and E are not on neighbouring chairs. Which one of the following pairs must be sitting on neighbouring chairs?

(a) A and B
(b) C and E
(c) B and F
(d) A and C

Answer(b) C and E

The same round table, reasoned rather than counted. That item fixes who must sit beside whom under stated conditions; this one counts how many seatings satisfy a condition of the same kind.

CAPF_GAI_2022_Q412022Open & attempt →

There is a group of 5 people among which there is one couple. In how many ways can these 5 people be seated in a row having 5 chairs if the couple is to be seated next to each other?

(a) 24
(b) 48
(c) 60
(d) 120

Answer(b) 48

The identical group of five with the identical couple, seated in a row instead of a circle and with the condition reversed. Comparing the two answers, 48 out of 120 in a row against 12 out of 24 in a circle, shows exactly what the circle costs.

Practice

practice — not a real PYQ
In how many ways can 6 people be seated around a circular table?
- (a)60
- (b)120
- (c)360
- (d)720
Answer(b) 120 — (6 − 1)! = 5! = 120.
practice — not a real PYQ
In how many ways can 5 people be seated in a row so that two particular people sit together?
- (a)24
- (b)48
- (c)72
- (d)120
Answer(b) 48 — four units give 4! = 24 orders, doubled for the internal swap.