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?

Correct — B, 48. Tie the couple together and treat the pair as one block. That leaves four objects to arrange in the row — the block plus the other three people — which can be done in 4! = 24 ways. Within the block the husband and wife can sit in either order, giving 2 arrangements, so the total is 24 × 2 = 48. The block method works whenever a set of people must stay adjacent.

• (a)24 — 24 is 4!, the count before the couple's own two orders are allowed for. Forgetting to multiply by 2 is the commonest slip in this question type.
• (c)60 — 60 matches no step of the calculation. It is half of 120 and would need the couple to be indistinguishable, which they are not.
• (d)120 — 120 is 5!, the number of ways to seat five people with no restriction at all. The adjacency condition must cut this down, not leave it untouched.

Arrangements with an 'always together' condition are handled by gluing the tied members into a single block, arranging the reduced set, and then multiplying by the internal arrangements of the block. The complementary condition, 'never together', is total minus together.

The check that keeps you honest is that the answer must be smaller than the unrestricted 120 and larger than the 24 you get if you forget the internal swap — and 48 is the only option in that window. Had the question asked for the couple never to sit together the answer would be 120 − 48 = 72. Had the five been seated around a circular table instead of in a row, the count would be 3! × 2 = 12, because a circle fixes one position.

• n distinct objects in a row: n! arrangements.
• If k specified objects must stay together, glue them: (n − k + 1)! × k!.
• Never together = total minus always together.
• Around a circle, n distinct objects give (n − 1)! arrangements, since rotations are not counted separately.

• Stopping at 4! and answering 24.
• Applying the circular formula to a row of chairs.
• Reading 'next to each other' as 'at the two ends', which is a different and much smaller count.

Asked as a seating or arrangement count with one adjacency condition, sometimes phrased negatively as 'never sit together'.

• practice — not a real PYQ

  In how many ways can 6 students stand in a row if two particular students must always stand together?

  • (a)120
  • (b)240
  • (c)360
  • (d)720

  Answer(b) 240 — glue the pair to get 5 objects, 5! = 120, and multiply by the pair's 2 internal orders.

• practice — not a real PYQ

  In how many ways can 5 people be seated around a circular table?

  • (a)24
  • (b)60
  • (c)120
  • (d)20

  Answer(a) 24 — circular arrangements of n objects number (n − 1)!, and 4! = 24.