Problem Solving

pemdas wrote:thanks for contributing, I guess the q. stem is ambiguous. I meant only women sit by two; you inferred two women sit together and the rest can be permuted. I tried to follow your explanations, you have seated two women next to each other and the other two women are ordered as such, in total three elements two women joint, one women and one women, 6P3 correct?

The only thing, I was concerned with why you have permuted here 6P3, don't you need the number of options for three seats? You are permuting among 4 (4P2) anyways, why you permute 6P3? Doesn't this add to the additional sets with other 6 men permuted already within the circular permutation (5!)? Would you reconsider, 6C3 instead?

rijul007 wrote:

pemdas wrote:question:
How can six men and four women sit at the table if only two women should sit next to each other? 5!*6C4*4P2

Permutation for 6 men => 5!

only two women should sit next to each other..
so first we select the two women who sit together == 4P2 ways
we take the two women as a single entity
so now we have 6 places for 3 ==> 6P3

So the no of arrangements should be 5!*6P3*4P2

Let us say V and X sit together
So i hv considered VX as a simple entity...
Now we need to arrange VX, Y and Z in 6 places...
No of ways to do that=> 6P3
arranging V and X among themselves => 2 ways
6P3*2 is the no of ways when V and X are together
We need to take into account other possible pairs too... 4C2
Total no of possible ways=> 5!*6P3*2*4C2

shekhar.kataria wrote:

LalaB wrote:A group of ten people (six men and four women) wants to sit in a circular table, and no 2 women want to seat together. in how many different ways can they seat down?

Dear Experts/ fellow members

Below is the method i approached, but the answer is not the same. I am sure i am counting some extra cases here but couldnt figure out which one. So here goes my attempt at it...

We have M M M M M M W W W W

Total number of ways of arranging 10 people on circular table --- 10-1= 9! ways

Now if we consider the case where all women will sit together and minus it from the total number of ways ,, that should be equal to saying that no 2 women sit together.

So for this we consider W W W W as one person and now we have M M M M M M WWWW.
These 7 can be arranged in circle in 6!*4!ways, 4!here signifies that all women can interchange their position.

So as per the logic total number of ways should be 9! - 6!*4! which is not equal to the correct answer above decided in majority.

Please correct me where i m going wrong. Does my way of doing it includes the cases where 2/3 women can sit together but not all women can sit together.????

Your method fails to include a lot of other cases in which women sit together. You also need to consider cases where the women are grouped as (3,1); (2,2); and (2,1,1).

I should warn you that you're heading down a long, tedious, difficult path if you pursue this by your method of joining the women together and treating them as one. For example, when you go to subtract out the cases where 3 women sit together and one woman is apart and you say let's treat the three women together as one unit and calculate the resulting circular permutation, that will also count cases where your lone woman ends up next to the group of 3, but this is really a group of 4, which was already subtracted out in your first calculation, so then you have to figure out how to account for that. And this will happen every time, and each one will get more complicated and aggravating to unravel. So if you decide to go this route, lots of luck to you.

If you want to pursue this by subtracting out the groups that include women sitting together, I would suggest finding the remaining ones along the lines of how we solved the original problem directly. That is, seat the men first in 5! ways, then figure out the spots for the women, and then permute the women.

So give it a shot if you want, but don't spend hours fighting through it if you find you're overmatched. As Mitch said, you almost surely won't encounter a problem like this on the real GMAT. So unless you're already at the 50-51 level, there's probably a better way for you to spend your prep time.

GmatMathPro wrote:Before I jump in here, let me make sure I understand. In the new question you're posing, you want the four women to be seated in pairs so that each woman is seated next to exactly one woman? Is that right?

yes, women sitting in pairs with each other

pemdas wrote:yes, women sitting in pairs with each other

Exactly one woman should be sitting adjacent to a woman

First we arrange all the men =>5!
Now make group of twos =>no of ways this can be done is 4C2
Each of the group has to be b/w any two men
You hace six plces for 2 pairs => 6P2
2 arrangements within each pair

so the total no of permutations would be => 5!*4C2*6P2*2*2

This is two times too big. Do you see why?

rijul007 wrote:

pemdas wrote:yes, women sitting in pairs with each other

Exactly one woman should be sitting adjacent to a woman

First we arrange all the men =>5!
Now make group of twos =>no of ways this can be done is 4C2
Each of the group has to be b/w any two men
You hace six plces for 2 pairs => 6P2
2 arrangements within each pair

so the total no of permutations would be => 5!*4C2*6P2*2*2

GmatMathPro wrote:This is two times too big. Do you see why?

rijul007 wrote:

pemdas wrote:yes, women sitting in pairs with each other

Exactly one woman should be sitting adjacent to a woman

First we arrange all the men =>5!
Now make group of twos =>no of ways this can be done is 4C2
Each of the group has to be b/w any two men
You hace six plces for 2 pairs => 6P2
2 arrangements within each pair

so the total no of permutations would be => 5!*4C2*6P2*2*2

I multiplied this by 4 to show the arrangement within each pair

A B V X C D Y Z E F
A B X V C D Y Z E F
A B V X C D Z Y E F
A B X V C D Z Y E F


or is their something else that I'm not seeing...???

that's what I was going to point too. I agree about 5!*6C4*4P2

GmatMathPro wrote:This is two times too big. Do you see why?

rijul007 wrote:

pemdas wrote:yes, women sitting in pairs with each other

Exactly one woman should be sitting adjacent to a woman

First we arrange all the men =>5!
Now make group of twos =>no of ways this can be done is 4C2
Each of the group has to be b/w any two men
You hace six plces for 2 pairs => 6P2
2 arrangements within each pair

so the total no of permutations would be => 5!*4C2*6P2*2*2

rijul007 wrote:

pemdas wrote:yes, women sitting in pairs with each other

Exactly one woman should be sitting adjacent to a woman

First we arrange all the men =>5!
Now make group of twos =>no of ways this can be done is 4C2
Each of the group has to be b/w any two men
You hace six plces for 2 pairs => 6P2
2 arrangements within each pair

so the total no of permutations would be => 5!*4C2*6P2*2*2

The problem is the statement in bold above. If you have 4 women, let's call them ABCD, then if you do 4C2, I'm assuming your idea is you choose 2 to make one group and then the leftover two is the other group. But this causes repeats. Observe:

ABCD

AB CD
AC BD
AD BC
BC AD
BD AC
CD AB

The bottom 3 are identical to the top 3. So there are really only 3 ways to break a group of 4 into groups of 2. This, incidentally, is the same reason your approach to the problem where the group of 6 is broken up into two groups of three didn't work. There are actually only 10 ways to break up a group of 6 into groups of 3, not 20.

Quant Breaker wrote:that's what I was going to point too. I agree about 5!*6C4*4P2

Interestingly enough, this is two times too small.

Here is what I would do.

1. Place the men: 5! ways.

2. Choose the two spots for the women: 6C2

3. Arrange the women among the 4 chairs: 4!

Putting it all together: 5!4!*6C2, which, bizarrely enough, is equivalent to the answer to the original question: 5!*6P4

When you say 6C4 and I put 6C2, it's the same as one is arranged around the two people and the other around four people leaving the same number of combination sets.
Where we differ is you put 4! and I have 4P2. You mentioned that two spots are chosen for the women, 6C2, right? Then why you have permuted 4P4 individually when you combine (or stick/clue) two women as one. Don't you need to permute 4P2 instead of 4P4?

GmatMathPro wrote:

Quant Breaker wrote:that's what I was going to point too. I agree about 5!*6C4*4P2

Interestingly enough, this is two times too small.

Here is what I would do.

1. Place the men: 5! ways.

2. Choose the two spots for the women: 6C2

3. Arrange the women among the 4 chairs: 4!

Putting it all together: 5!4!*6C2, which, bizarrely enough, is equivalent to the answer to the original question: 5!*6P4

I'm not combining the two women as one. You have the 6 men seated around the table. Then you pick the two spots for the women. Imagine 2 chairs are then brought to each of these two spots. Number the chairs 1,2,3,4. We can assign the women to the chairs in 4! ways.

two chairs brought requires multiplication by 2 (x2) and then you extend the other sets as well. If you are using the table from the previous page 4 men and 2 women - it works, but not for six men and four women. It cannot be 5!*6P4 again with six men and four women

two spots for the women whose number four means you stick two women as one and assign one option (place) for two of them. Within this place two women may permute (change) their order as AB and BA.

GmatMathPro wrote:I'm not combining the two women as one. You have the 6 men seated around the table. Then you pick the two spots for the women. Imagine 2 chairs are then brought to each of these two spots. Number the chairs 1,2,3,4. We can assign the women to the chairs in 4! ways.

Okay then, consider this. Let's calculate the answer to the original problem (no two women can sit together) by taking the total ways to arrange 10 people around the table and subtracting the cases where at least two women are together. First, I'll calculate number of ways we could arrange the people so that at least two women are sitting together using the method I'm advocating:

1. All four women are sitting together:

Arrange the men: 5!
Choose a spot for the women: 6
Arrange the women: 4!

Total: 6*5!*4!

2. Three women sit together and one woman sits by herself:

Arrange the men: 5!
Choose a spot for the group of three women: 6
Choose a spot for the lone woman: 5
Arrange the women: 4!

Total: 6*5*5!*4!

3. There are two separate pairs of women (the problem we're currently debating):

Arrange the men: 5!
Choose two spots for the pairs of women: 6C2=15
Arrange the women: 4!

Total: 15*5!*4!

4. There is one pair of women and the other two women sit by themselves:

Arrange the men: 5!
Choose a spot for the pair:6
Choose the two spots for the lone women: 5C2=10
Arrange the women: 4!

Total: 10*6*5!4!

Add them all up: 6*5!4!+6*5*5!4!+15*5!4!+10*6*5!4!=

5!4!(6+30+15+60)=
111*5!4! ways to seat 6 men and 4 women so that at least two women are sitting together.

Now, 10 people can be seated around a circular table in (10-1)!=9! ways. So, the number of ways to seat them so that no two women sit together would be 9!-111*5!4! which equals 43,200 or 5!*6P4 which you can verify with a calculator or by the following method:

9!-111*5!4!=

5!(9*8*7*6-111*4!)=
5!((3*3)(2*2*2)(7)(2*3)-3*37*(2*2)(3)(2))=
5!(2*2*2*2*3*3*3*7-2*2*2*3*3*37)=
5!(2*2*2*3*3[2*3*7-37])=
5!(2*2*2*3*3[42-37])=
5!(2*2*2*3*3*5)=
5!(6*5*4*3)=
5!*6P4

So, 9!-111*5!4!=5!*6P4, which agrees with the answer we got when we calculated it directly. So, 111*5!*4! must be the correct answer to the number of ways we can arrange the people with at least some women together. So either the components I calculated above are correct, or they contain a bunch of errors which all miraculously cancel out to give me the right answer. However, I used the same method for all of the components (place the men, pick the spots for the women, permute the women) for every component, so if this were wrong, it should either consistently overestimate or consistently underestimate the true numbers. But if this is the case there's no way the errors would cancel out. Therefore, I submit that the method is correct.

pemdas wrote:two chairs brought requires multiplication by 2 (x2) and then you extend the other sets as well. If you are using the table from the previous page 4 men and 2 women - it works, but not for six men and four women. It cannot be 5!*6P4 again with six men and four women

two spots for the women whose number four means you stick two women as one and assign one option (place) for two of them. Within this place two women may permute (change) their order as AB and BA.

GmatMathPro wrote:I'm not combining the two women as one. You have the 6 men seated around the table. Then you pick the two spots for the women. Imagine 2 chairs are then brought to each of these two spots. Number the chairs 1,2,3,4. We can assign the women to the chairs in 4! ways.

GmatMathPro wrote:

rijul007 wrote:

pemdas wrote:yes, women sitting in pairs with each other

Exactly one woman should be sitting adjacent to a woman

First we arrange all the men =>5!
Now make group of twos =>no of ways this can be done is 4C2
Each of the group has to be b/w any two men
You hace six plces for 2 pairs => 6P2
2 arrangements within each pair

so the total no of permutations would be => 5!*4C2*6P2*2*2

The problem is the statement in bold above. If you have 4 women, let's call them ABCD, then if you do 4C2, I'm assuming your idea is you choose 2 to make one group and then the leftover two is the other group. But this causes repeats. Observe:

ABCD

AB CD
AC BD
AD BC
BC AD
BD AC
CD AB

The bottom 3 are identical to the top 3. So there are really only 3 ways to break a group of 4 into groups of 2. This, incidentally, is the same reason your approach to the problem where the group of 6 is broken up into two groups of three didn't work. There are actually only 10 ways to break up a group of 6 into groups of 3, not 20.

Oh yes, i should have seen that...
Thanks..

Pete, I understood that I was applying additional (unnecessary) restriction as 4P2, instead I could distribute freely 4! or leave three women for distribution with the remaining seats 4P3. Thanks for taking your time and shedding light to this q.