Find the number of ways in which four men, two women and a

Note:
The number of arrangements of n distinct objects in a row is given by \(n!\).
The number of arrangements of n distinct objects in a circle is given by \((n-1)!\).

We have M, M, M, M, W, W, C --> glue two women and the child so that they become one unit and the child is between women: {WCW}. Now, these 5 units: {M}, {M}, {M}, {M}, {WCW} can be arranged around the table in (5-1)!=4! ways and the women within their unit can be arranged in 2 ways {W1, C, W2} or {W2, C, W1} so total # of arrangement is 4!*2=48.

Answer: 48.

240 would be the answer if the arrangement were in a row: {M}, {M}, {M}, {M}, {WCW} can be arranged in a row in 5! ways and the women within their unit can be arranged in 2 ways {W1, C, W2} or {W2, C, W1} so total # of arrangement is 5!*2=240.

P.S. Please read and follow: how-to-improve-the-forum-search-function-for-others-99451.html

So please provide answer choices for PS questions.
_________________

Hi

Thanks for your reply. And I did not have the answer, so could not post it, apologies.

One question, if the places in table are numbered, wouldn't it become a case like arranging the members in the stated manner in a line, in which case the answer 240 is valid ?

Regards,
Subhash
_________________

If the chairs are numbered and one specific arrangement and the same arrangement but shifted by one position are considered different then the answer will simply be 48*7.

That's because the difference between placement in a row and that in a circle is following: if we shift all object by one position, we will get different arrangement in a row but the same relative arrangement in a circle.
_________________

For this szenario I received 744 different cases, including the cases when the child and a man or men are between the 2 women.

{M}, {M}, {M},{WCMW} 4*3*2*2*2=96
{M}, {M},{WCMMW} 3*2*3*2=72
{M},{WCMMMW}2*4*3*2*2=96
{WCMMMMW}5*4*3*2*2=240
These possibilities together with Bunuels possibilities when only the child is btw the women {WCW} gives 504+240=744

Please correct me if I am wrong!

I think the question means that ONLY the child must be between two women.
_________________

Ok, I see.
If the question were stated for a row and not a table, and if there was the ONLY requirement, would that be correct?

If... If... This question is already not a GMAT type...

But anyway your solution is still wrong:

{M}, {M}, {M}, {M}, {WCW}: 5!*2=240;
{M}, {M}, {M}, {WCMW}: 4!*4C1*2!*2;
{M}, {M}, {WCMMW}: 3!*4C2*3!*2;
{M}, {WCMMMW}: 2!*4C3*4!*2;
{WCMMMMW}: 5!*2.
_________________

Another great explanation. Thank you Bunuel

Another way to look at the question.
(though let me point out first that the question doesn't specifically say 'circular table'. It just says 'sit at a table'. If it is a rectangular table, perhaps there are 4 chairs on one side, 3 on the other etc. Since you mentioned "Circular Permutation Problem" in the subject line, I am assuming it is meant to be a circular table.)

7 seats around a circular table, 7 people.
First I make the child sit anywhere in 1 way since all seats are the same. The two women can sit around him in 2! ways. Now 4 seats are left for 4 men and they can occupy them in 4! ways.
Total number of ways = 4!*2! = 48

Another thing, if the places are numbered, say 1, 2, 3 etc for the 7 seats, the number of arrangements will be 7*2!*4! = 336.
Make the child sit on any one of the 7 seats since all are unique now. The women sit around the child in 2! ways and the men sit on the rest of the 4 seats in 4! ways.
The reason why this number is greater than the number of arrangements in case of a row (240 ways) is because in a row, child cannot be in 1st or 7th position while in a circle, the child can sit on seat no 1 or seat no 7. So we have 2*2!*4! = 96 extra cases in case of numbered seats around a circular table.
Note: 240 + 96 = 336
_________________

Treat the women and child as a unit.

W C W

So in total there are (5-1)! = 24 ways for arrangement

24 x 2 = 48 <---- Women can swap places with respect to the child

Answer is 48.
_________________

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________