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

Answer 1

Answer: (A)

$5\times 6!$

Number of ways of arranging $5$ boys and $3$ girls, i.e. $8$ people on a round table would be $7!$

We subtract the number of ways of arranging those people when $B_1$ and $G_1$ are always together to get the required answer.

When $B_1$ and $G_1$ are together, we get $4$ boys $+$ $2$ girls $+1(B_1+G_1)$ i.e. $7$ people and since $B_1+G_1$ can be permuted in $2$ ways, these can be arranged in $6!\times 2$ ways.

Subtracting, we have $7!-6!\times 2=6!(7-2)=5\times 6!$ ways in total