Question

4 boys and 7 girls are to sit in a row at random.The probability that no two boys will sit together is

• A. $\frac{7!\times 4!}{11!}$
• B. $\frac{7!{\times }^8{P}_4}{11!}$
• C. $\frac{7!{\times }^8{C}_4}{11!}$
• D. $\frac{7!\times 8!}{11!}$

Answer 1

The correct option is B $\frac{7!{\times }^8{P}_4}{11!}$

The condition is that no two boys sit together.

We start by arranging the girls. $7$ girls can be arranged in $7!$ ways.

Once the girls are arranged , a boy can occupy a position between any two girls or the leftmost or rightmost position as shown in figure.

Hence, out of $8$ positions, $4$ boys can be arranged in any $4$ positions.

The number of ways of arranging $4$ boys in available $8$ positions is ${}^8{P}_4$
Hence, total number of arrangements such that no two boys sit together is $7!{\times }^8{P}_4$

Without any restriction, the $11$ guys can be arranged in $11!$ ways.

Hence, probability
$=\frac{7!{\times }^8{P}_4}{11!}$