Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

now consider one position between each two adjecent men. 4 women need to sit in these 4 spots in order for condition to be valid (no two men together). this can, again, be done in:
4P4 = 24 ways.

in order to have 4 women and 4 men together at a table without adjecent men, there is 1 general layout but in reality 2 configuration : MWMWMWMW or WMWMWMWM

4 positions for men and they are 4 so : 4! ways (=24)
4 positions for women and they are 4 so : 4! ways (=24)

Take this problem as an example: Arrange 3 men and 3 women to a round table so that no two men are adjacent.

There are 2*3! way to arrange the men to the table (to 1st, 3rd and 5th AND to 2nd, 4th and 6th) (let's call these 3 men A, B and C)
After arranging the men, there are 3! way to arrange the women to the table (call them 1, 2 and 3)

However, this is a round table, in which the arrangement A1B2C3 happens 6 times in 2*3!*3! of arrangements. Similar to other arrangement.

Therefore, the answer for my example is 2*3!*3!/6 = 3!*3!/3 = 3!*(3-1)!

Similarly, the answer for the posted question is 4!*(4-1)!

However, this is a round table, in which the arrangement A1B2C3 happens 6 times in 2*3!*3! of arrangements. Similar to other arrangement.

Therefore, the answer for my example is 2*3!*3!/6 = 3!*3!/3 = 3!*(3-1)!

Sorry, I still try to get this one
We are talking about arragement , right ? It should be permutations no ? I don't understand why you need to divide by 6 ?
For me the total number of possibilities is 2*3!*3!
If someone can explain please