Problem Solving

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?

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?

Arrange the 6 men in circular fashion in (6-1)! ways
now we have 6 alternative spaces between men, of which if we select any four places and arrange four women, then given condition will get satisfied. => 6P4

IMO it is 5!*6p4.

user123321

user123321 wrote: Arrange the 6 men in circular fashion in (6-1)! ways
now we have 6 alternative spaces between men, of which if we select any four places and arrange four women, then given condition will get satisfied. => 6P4

IMO it is 5!*6p4.

user123321

thanks for sharing ur thoughts )

here is what I think-
1st we arrange men in (6-1)! ways.
then we have 4 women, 6 spaces between men, and no two women want to seat together.

so we arrange women in 6C4* 4! ways

as a result we have - 5!*4!*6C4

lets find the truth )

Hey, My math is crap
(6-5)! is fine for me
I could not understand the logic of multiplying 6C4 with 4 again.

Lala B, could you throw some light?

gunjan1208 wrote:Hey, My math is crap
(6-5)! is fine for me
I could not understand the logic of multiplying 6C4 with 4 again.

Lala B, could you throw some light?

When 6 men are arranged in a circle, we will have 6 spaces in between them.
if we arrange the women in those spaces, then no two women will come next to each other.
so we have 6 spaces & 4 women to arrange in them = 6p4 ways.
or else
you can select 4 spaces out of 6 spaces and arrange four women in them = 6c4 * 4! = 6p4 (both mean the same)

user123321

I would disagree as the found arrangement seems mechanized
@user, why you would arbitrarily assign "1 man=empty space=1 man"? I understand if you have #men=#women, but here it's different (unlike similar q. discussed on gmat-club), the logic isn't followed here

there can be well arranged the order "1 man=1 man=empty space" in the beginning, middle and the end of the circle.

IOM you overestimated the number of possible ways by 3*5!

here's my take on this q.
>> more slots with say man-man-woman-man ... in the beginning, middle, end. Therefore, i would agree about cluing two men and permuting them through four slots available for women. Final order could be (6-1)!*4!*4P2=(5!*4!)*12

Hi Pemdas

pemdas wrote: there can be well arranged the order "1 man=1 man=empty space" in the beginning, middle and the end of the circle.

here empty space doesn't mean it is a physical blank space, it just implies whether there is a possibility for a women to come next to that man or not.

pemdas wrote: here's my take on this q.
>> more slots with say man-man-woman-man ... in the beginning, middle, end. Therefore, i would agree about cluing two men and permuting them through four slots available for women. Final order could be (6-1)!*4!*4P2=(5!*4!)*12

I am assuming cluing means combining, if you combine two men like that, you will miss the permutations where a women can happen to sit in between them

Anyways could you just elaborate your solution, if i misunderstood it?

Thanks
user123321

user, when you set 6P4 you make prearranged set of 4 elements (women) through 6 options (seats). What I'm asking is why you set 6 options when it can be 4 minimum possible options. If the number of men were equal to the number of women - say 6 women, then you had to allow for 6 options (seats) at least, because with nPk, n >=k is the only available range. Here we have 4 women, hence when you assign 6P4 you do this arbitrarily.

When I set 4P2 I don't miss any women in between the men, because women are less than men by 2 and in any way we can't have more than 4 women to be placed between 4 men, the residual is 2 men. That's why I arrived to the solution (5!*4!)*4P2

please let me know your opinion on that

first arrange 6 men on the circular table => 5! = 120

No two women should be together..
there can be not more than 1 women b/w two men

In total there are 6 places you can choose from
for 4 women.. the no of selections would be 6C4 = 15

Total no of arrangements = 5!*15*4! or 5!*6P4

Can we have expert help here, please! I dont know what is definitely right.

GMAT Pro, plz.

rijul007 wrote:there can be not more than 1 women b/w two men

why there should be 'b/w two men', can't this be among three or four men? Men exceed women by two, hence one women can be placed such as 'three men to the left-one woman-one man to the right', 'two men to the left-one woman-two men to the right', etc.

In total there are 6 places you can choose from
for 4 women.. the no of selections would be 6C4 = 15

There can be 6 or 5 or 4 places to choose from ... When we say 6 places, we prearrange the set of options available for women. We know that at least there are four places, since four women must be allocated anyways, hence 4!(or 4P4). We also know that two additional men offering two more places can be permuted among (through, within) four women or four places occupied by women such as 4P2. Total makes (6-1)!*4!*4P2

pemdas wrote:

rijul007 wrote:there can be not more than 1 women b/w two men

why there should be 'b/w two men', can't this be among three or four men?

there can be not more than 1 women b/w two men

no of women b/w any two men can be either 0 or 1

pemdas wrote:

rijul007 wrote:there can be not more than 1 women b/w two men

why there should be 'b/w two men', can't this be among three or four men? Men exceed women by two, hence one women can be placed such as 'three men to the left-one woman-one man to the right', 'two men to the left-one woman-two men to the right', etc.

In total there are 6 places you can choose from
for 4 women.. the no of selections would be 6C4 = 15

There can be 6 or 5 or 4 places to choose from ... When we say 6 places, we prearrange the set of options available for women. We know that at least there are four places, since four women must be allocated anyways, hence 4!(or 4P4). We also know that two additional men offering two more places can be permuted among (through, within) four women or four places occupied by women such as 4P2. Total makes (6-1)!*4!*4P2

I hope this image makes it clearer

thanx for draw.
fine, you first sit men and then you sit women between men

let's assume A B C D E F are different men and V X Y Z are women. When six men are seated there will be six slots available for women but women will occupy only four slots. For every ordered arrangement of six men you take 6P4 or simply said 360 ordered arrangements of women. Now my question is 'if A B C D E F are permuted, ordered 5! times (since circ. perm, due to clock and counter wise direction changes starting from the initial point) for every of 5! sets of men you assign 360 ordered arrangements of women? Your set A B C D E F is unique as one of total 5! arranged sets, and the assignment of woman set such as A-V B-X C-Y D-Z E F is the same as F E D-Z C-Y B-X A-V in circular order. If you had six women and six men then it wouldn't be like this, because all six possible slots between men were taken by women, but here it's repeating the same order of men-women. Therefore you over-count by six times each of the unique sets (ABCDEF), i.e. you assign six times more to each of 5! man sets.

Therefore, I first sit men like you do, then sit women in any of the four available slots (unlike you do) 5!*4! and for every combination of 5!*4! I'm taking additional 12 arrangements of two men reseated after six men and four women were seated. These two men are seated by each woman in different orders (they are permuted) A B C D E F -> A-V B-X C-Y D-Z E F -> A E F -V B-X C-Y D-Z -> A F E -V B-X C-Y D-Z E F. Please note when the order is turned backwards (counter-clock wise) I have not the identical orders, but for example two new ones -> F E A-V B-X C-Y D-Z and E F A-V B-X C-Y D-Z

5!*4!*4P2 against your order 5!*6P4, the difference is (5!*6) less than yours

pemdas wrote:thanx for draw.
fine, you first sit men and then you sit women between men

let's assume A B C D E F are different men and V X Y Z are women. When six men are seated there will be six slots available for women but women will occupy only four slots. For every ordered arrangement of six men you take 6P4 or simply said 360 ordered arrangements of women. Now my question is 'if A B C D E F are permuted, ordered 5! times (since circ. perm, due to clock and counter wise direction changes starting from the initial point) for every of 5! sets of men you assign 360 ordered arrangements of women? Your set A B C D E F is unique as one of total 5! arranged sets, and the assignment of woman set such as A-V B-X C-Y D-Z E F is the same as F E D-Z C-Y B-X A-V in circular order. If you had six women and six men then it wouldn't be like this, because all six possible slots between men were taken by women, but here it's repeating the same order of men-women. Therefore you over-count by six times each of the unique sets (ABCDEF), i.e. you assign six times more to each of 5! man sets.

i didnt get you here... can you please eloborate on the highlighted part