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

Answer 1

The Correct Option is D

First, consider arranging the five boys around the round table. Since this is a circular arrangement, the number of ways to arrange them is given by \((n-1)!\), where \(n\) is the number of boys.

Therefore, the number of ways to arrange the boys is \(4! = 24\).

Once the boys are seated, they form \(5\) spaces in which the girls can be seated such that no two girls are together.

We now need to arrange \(3\) girls in these \(5\) spaces. The number of ways to choose \(3\) spaces out of \(5\) is represented by the combination formula:

\[\binom{5}{3} = 10\]

For each of these space combinations, the \(3\) girls can be permuted among themselves in \(3!\) ways.

Thus, the number of ways to sit the girls is given by:

\(10 \times 3! = 10 \times 6 = 60\)

Combining these, the total number of arrangements where no two girls sit together is:

\(24 \times 60 = 1440\)

Therefore, the number of ways the boys and girls can sit around the table so that no two girls are together is 1440.