In how many ways can $7$ men and $7$ women can be seated around a round table such that no two women can sit together?

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

${(7!)}^{2}$

$7! \times 6!$

${(6!)}^{2}$

$7!$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Given : $7$ men and $7$ women seated around a round table

To find: no of ways in which $7$ men and $7$ women can be seated around a round table such that no two women can sit together.

A permutation is a (possible) rearrangement of objects.

By using formula $n! = 1 . 2 . 3 . . . . . . . . (n - 1) . n$

$7$ men can seat themselves in a round table in $(7 - 1)! = 6!$ ways .

Now, there are $7$ places vacant between these $7$ men.

$∴ 7$ women can seat themselves in these $7$ places in $7!$ ways.

$∴$ Total number of required arrangement where no two women sit together $= 6! \times 7!$

Watch 3-min video & get full concept clarity