Question

Five men, two women, and a child sit around a table. Find the number of arrangements where the child is seated between two men

Answer 1

Here, 5 men + 2 women + 1 child = 8 people.

Child is seated between two men:

Two men can be selected from the five men (M₁, M₂, M₃, M₄, M₅) in 10 ways as follows:

(M₁M₂, M₁M₃, M₁M₄, M₁M₅, M₂M₃, M₂M₄, M₂M₅, M₃M₄, M₃M₅, M₄M₅)

Consider 2 men and a child as one group. These two men can be arranged so that the child is in between the men, in ²P₂ = 2! ways.

Now the group of 2 men and a child, the remaining 3 men and 2 women, i.e., altogether 6 persons can be arranged around the table in (6 – 1)! = 5! ways.

∴ the number of ways of arranging eight people such that the child is seated between two men

= 10 × 2! × 5!

= 10 × 2 × 1 × 5 × 4 × 3 × 2 × 1

= 2400