Question

The number of ways in which 5 boys and 3 girls can be seated on a round table, if a particular boy $B_1$ and a particular girl $G_1$ never sit adjacent to each other, is

• A. 7!
• B. 5 $\times$ 6!
• C. 6 $\times$ 6!
• D. 5 $\times$ 7!

Answer 1

Answer: (B)

5 $\times$ 6!

Answer 2

1. Total arrangements without restriction:
   Since seating is around a round table, the total number of arrangements for 8 persons is

   $$(8-1)! = 7!$$

2. Arrangements where $B_1$ and $G_1$ are adjacent:
   Fix $B_1$ (to account for rotational symmetry). Then, $G_1$ can be seated in 2 positions (the two seats adjacent to $B_1$). The remaining 6 persons can be arranged in $6!$ ways.

   $$\text{Adjacent arrangements} = 2 \times 6!$$

3. Arrangements where $B_1$ and $G_1$ are not adjacent:

   $$7! - 2 \times 6!$$

   Factor $6!$ out:

   $$= 6! \times (7 - 2) = 6! \times 5 = 5 \times 6!$$

Total ways = $7!$. Ways with $B_1, G_1$ adjacent = $2 \times 6!$. Thus, ways with them non-adjacent = $7! - 2 \times 6! = 5 \times 6!$.