4 Men - 4 Women

[qhoc0010]

How many ways can 4 men and 4 women sit at a round table with no two women in consecutive positions?

(A) 24

(B) 72

© 288

(D) 144

(E) 48

[GmatGnat]

Answer (D)

 

In order to have no two women in consecutive positions, we must alternate between men and women. So we fix a woman in one position. The other women can be arranged 3! ways (since there are 3 of them left). Between the women, the men can be arranged 4! ways (since there are 4 of them).

 

3! * 4! = 144

Answer (D)

[qhoc0010]

OK, thanks

 

But what is the general way to approach these "round table" problem?

Let say:

n women and n men

(n-1)! x (n)! --> CORRECT?

 

But what if:

n women, n men, n lesbians (sorry)

How to solve this?

[thebullfighter]

OK, thanks

 

But what is the general way to approach these "round table" problem?

Let say:

n women and n men

(n-1)! x (n)! --> CORRECT?

 

But what if:

n women, n men, n lesbians (sorry)

How to solve this?

Not Correct. It's (2n-1)!

 

for the 2nd Q, the lesbians one. Are the same n women Lesbians or are these 2 categories diff. ie. Are there a total of 3n people?? if they are diff. Then answer is (3n-1)!

 

in general, for n people to be seated around a round table, (n-1)! ways.

[qhoc0010]

How come it is (2n -1)! ? The initial question has : 3! x 4!

according to the formular: (2x4 - 1)! = 7! ????

 

For my second question: yes, all diff.

[thebullfighter]

How many ways can 4 men and 4 women sit at a round table with no two women in consecutive positions?

the initial Q has a condition. :)

 

But what is the general way to approach these "round table" problem?

Let say:

n women and n men

(n-1)! x (n)! --> CORRECT?

for this Q, Total= n+n= 2n

 

For round table formula is: (Total-1)!

hence, (2n-1)!

 

similarly for lesbian Q, Total=3n, Hence, (3n-1)!

[qhoc0010]

I mean just like the Q condition in the initial.

 

How many ways can "n" men and "n" women sit at a round table with no two women in consecutive positions?

 

How many ways can "n" A, "n" B, and "n" C sit at a round table without no two A, no two B, or no two C in consecutive positions?

[thebullfighter]

Ohh!! sorry dear, then it's correct.

 

(n-1) ! * n!

 

same logic goes with the 2nd part.

[GmatGnat]

So how would we answer the second part of ghoc0010's question?

 

How can we arrange 5 lions, 5 tigers, and 5 bears around a table without seating like animals next to each other?

 

Is it (5-1)! * 5! * 5! = 4! * 5! * 5! = 345,600 ?

[thebullfighter]

just an additional *2 with it.

 

#ABCABCABCABCABCABC#

#ACBACBACBACBACBACB#

[qhoc0010]

It is getting complicated. Where did you get the " times 2"?

 

What if the problem like this:

How can we arrange 5 lions, 5 tigers, 5 bears, 5 chickens, 5 monkeys around a table without seating like animals next to each other?

[GmatGnat]

I think I see. The *2 comes from the fact that once we fix our first lion, we can then start with a tiger, or with a bear.

 

So we could arrange 5 lions, 5 tigers, 5 bears, 5 chickens, and 5 monkeys around a table without seating like animals next to each other in this many ways:

 

(5-1)! * 5! * 5! * 5! * 5! * 4!

4! * 5! * 5! * 5! * 5! * 4!

 

The first 4! comes from fixing our first lion, then we have 4 more lions to distribute. The next four 5!s come from distributing our tigers, bears, chickens, and monkeys around the table. The last 4! comes from the fact that we have 4! ways of ordering the tigers, bears, chickens, and monkeys.

 

How'd I do, bull?

[thebullfighter]

Perfect GG. :)