Arrange the 6 men in circular fashion in (6-1)! waysLalaB 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?
thanks for sharing ur thoughts )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
When 6 men are arranged in a circle, we will have 6 spaces in between them.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?
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: there can be well arranged the order "1 man=1 man=empty space" in the beginning, middle and the end of the circle.
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 thempemdas 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
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.rijul007 wrote:there can be not more than 1 women b/w two men
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!*4P2In total there are 6 places you can choose from
for 4 women.. the no of selections would be 6C4 = 15
pemdas wrote:why there should be 'b/w two men', can't this be among three or four men?rijul007 wrote:there can be not more than 1 women b/w two men
pemdas wrote: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.rijul007 wrote:there can be not more than 1 women b/w two menThere 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!*4P2In total there are 6 places you can choose from
for 4 women.. the no of selections would be 6C4 = 15
i didnt get you here... can you please eloborate on the highlighted partpemdas 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.
