The number of ways, in which 5 girls and 7 boys can be seated at a round table so that no …

Question

The number of ways, in which 5 girls and 7 boys can be seated at a round table so that no two girls sit together, is :

Accepted Answer

We have,

Number of girls $=5$

Number of boys $=7$

So, number of ways of arranging boys

around the table $=6$ ! and 5 girls can be arranged in 7 gaps in ${ }^7 \mathrm{P}_5$ ways.

$	herefore$ Required no. of ways $=6 ! 	imes{ }^7 \mathrm{P}_5$ $=126 	imes(5 !)^2$

Solutions · 1