Question

There are five boys and three girls who are sitting together to discuss a management problem at a round table. In how many ways can they sit around the table so that no two girls are together?

• A. 1220
• B. 1400
• C. 1420
• D. 1440

Hint:

Seat the boys first in a circle, then place the girls one per gap between boys so no two girls are adjacent.

Answer 1

The Correct Option is D

An equivalent way to count the girls' placement is to directly count ordered selections of gaps using a permutation.

1. Seat the boys in a circle. Five boys around a round table can be arranged in $(5-1)! = 4! = 24$ ways, and this creates exactly 5 gaps between them.
2. Place the girls using a permutation. To seat 3 girls into 3 of these 5 distinct gaps, with order mattering (since the girls are distinct people occupying distinct seats), we use $^5P_3 = \dfrac{5!}{(5-3)!} = \dfrac{120}{2} = 60$.
3. Multiply the two independent arrangements. Total ways $= 24 \times 60 = 1440$.

This matches the combination-then-factorial approach ($^5C_3 \times 3! = 10 \times 6 = 60$), confirming the answer is $1440$, option D.