$5$ boys and $5$ girls had to sit alternately around a round table. This can be done in $N$ ways. Then the value of $N$ is

Open in App

Solution

Five boys can be arranged in a circle in $4!$ ways.
After that, girls can be arranged in the five gaps shown as $‘ X^{'}$ in $5!$ ways.

Hence, total number of ways $= 4! \times 5! = 2880$

Suggest Corrections

Similar questions

5

boys and

5

girls had to sit alternately around a round table. This can be done in

N

ways. Then the value of

N

Q. In how many ways 8 boys and 5 girls can sit around a round table, so that no two girls sit together.

Q. 5 girls and 5 boys can be seated around a circular table such that no 2 girls will sit together in ways

5 girls and 5 boys are to be seated around a circular table such that no 2 girls will sit together. The number of ways in which they can be seated around the table is

Q. In how many ways

10

boys and

5

girls can sit around a circular table, so that no two girls sit together?

Join BYJU'S Learning Program

Explore more

Circular Permutation

Standard XII Mathematics

Join BYJU'S Learning Program

5 boys and 5 girls had to sit alternately around a round table. This can be done in N ways. Then the value of N is

Five boys can be arranged in a circle in 4! ways. After that, girls can be arranged in the five gaps shown as ‘X′ in 5! ways. Hence, total number of ways =4!×5!=2880

$5$ boys and $5$ girls had to sit alternately around a round table. This can be done in $N$ ways. Then the value of $N$ is

Five boys can be arranged in a circle in $4!$ ways.
After that, girls can be arranged in the five gaps shown as $‘ X^{'}$ in $5!$ ways.

Hence, total number of ways $= 4! \times 5! = 2880$