“m” men and “n” women are to be seated in a row so that no two women sit together. If m > n, then number of ways in which they can be seated, is

Correct — option (3), (|m |m + 1) / (|m – n + 1).

In the options, |m is an older way of printing m factorial, m!. Option (3) reads m! × (m + 1)! ÷ (m − n + 1)!.

Step 1 — seat the men first. The m men can be arranged in a row in m! ways.

Step 2 — count the gaps. A row of m men leaves m − 1 gaps between them and one gap at each end.
_ M _ M _ … _ M _
Gaps = (m − 1) + 2 = m + 1

Step 3 — place the women. At most one woman per gap keeps any two women apart. Giving each of the n women a different gap, in order:
ᵐ⁺¹Pₙ = (m + 1)! ÷ (m + 1 − n)!
The condition m > n makes sure there are enough gaps.

Step 4 — multiply the two stages.
Ways = m! × (m + 1)! ÷ (m − n + 1)! = (|m |m + 1) / (|m – n + 1)

Check with m = 3, n = 2: 3! × 4! ÷ 2! = 6 × 24 ÷ 2 = 72. Directly, 5 people sit in 5! = 120 ways; the two women sit together in 4! × 2! = 48 of them; 120 − 48 = 72.

The idea to remember: seat the people without a restriction first, then put the people who must be kept apart into the gaps.

Why the others are wrong

• (1)
  (|m |m – 1) / (|m – n + 1) — This is m! × (m − 1)! ÷ (m − n + 1)!. The part after m! is not the number of ways to place n women in the gaps.

  For m = 3 and n = 2 it gives 6 × 2 ÷ 2 = 6, while the actual count of seatings with the two women apart is 72, as 120 − 48 = 72 confirms.

• (2)
  (|m – 1 |m + 1) / (|n |n – 1) — This is (m − 1)! × (m + 1)! ÷ [n! × (n − 1)!]. Its first factor, (m − 1)!, is the number of ways to seat m people around a round table, but the stem seats everyone in a row, where m men can be arranged in m! ways.

  For m = 3 and n = 2 the formula gives 2 × 24 ÷ 2 = 24, not 72.

• (4)
  (|m |m – 1) / (|m – n – 1) — This is m! × (m − 1)! ÷ (m − n − 1)! = m! × ᵐ⁻¹Pₙ. It uses only the m − 1 gaps between the men, so it counts only the seatings in which no woman sits at either end of the row.

  The stem only forbids two women sitting together, so the end seats are allowed. For m = 3 and n = 2 this gives 12, not 72.

The multiplication principle: if one stage can be done in a ways and the next in b ways, both together can be done in a × b ways.

n! = 1 × 2 × … × n, with 0! = 1. The number of ways to choose r of n distinct places in order is ⁿPᵣ = n! ÷ (n − r)!.

The gap method keeps people apart. Arrange everyone else first; the spaces between them and at the two ends are the only seats where the separated people can go without touching one another.

Key facts

• n! = 1 × 2 × … × n and 0! = 1; some older books print n! as |n.
• ⁿPᵣ = n! ÷ (n − r)! counts ordered choices of r places from n.
• A row of m people has m + 1 gaps: m − 1 between them and one at each end.
• Row seatings of m men and n women with no two women together: m! × ᵐ⁺¹Pₙ = m!(m + 1)!/(m − n + 1)!.
• If no woman may sit at an end either, only the m − 1 inner gaps remain: m! × ᵐ⁻¹Pₙ.

Common traps

• Counting only the m − 1 gaps between the men. The two end seats are also open to a woman, so there are m + 1 gaps.
• Choosing the gaps with ᵐ⁺¹Cₙ and stopping. The women are different people, so which woman takes which gap matters: ᵐ⁺¹Cₙ × n! = ᵐ⁺¹Pₙ.
• Subtracting only the seatings where all the women form one block. With three or more women, that still leaves seatings where two of them sit side by side.

• practice — not a real PYQ

  In how many ways can 4 boys and 3 girls sit in a row so that no two girls sit together?

  • (a)144
  • (b)1,440
  • (c)240
  • (d)5,040
  Answer(2) — Seat the boys in 4! = 24 ways, leaving 5 gaps; place the 3 girls in ⁵P₃ = 60 ways; 24 × 60 = 1,440. Option (1) uses only the 3 gaps between boys; option (3) uses ⁵C₃ = 10 and ignores which girl sits where; option (4) is 7!, with no restriction.
• practice — not a real PYQ

  In how many ways can 3 men and 2 women sit in a row so that the two women sit together?

  • (a)24
  • (b)48
  • (c)72
  • (d)120
  Answer(2) — Treat the two women as one block: 4 units in 4! = 24 ways, and the women swap within the block in 2! = 2 ways; 24 × 2 = 48. Option (1) forgets the swap; option (3) counts seatings with the women apart; option (4) is all 5! seatings.