perms/combs question (1 Viewer)

[trecex1]

In how many ways can 4 men and 4 women be arranged around a circular table if:
All men are in pairs separated by two pairs of women.

Leave some reasoning in the steps pls, ty.

 

[leehuan]

This question is just confusing me... separated by "two pairs of women".

Unless you mean a pair of women on each side. Are you saying something like this?


M M F F M M F F (assume that this straight line is a circle)

 

[trecex1]

Its from the ruse 2012 trial, answer is 288. I don't understand their working at all. But that's the question word for word, i think it is how you drew it.

 

[leehuan]

I have a feeling I may be wrong. Someone correct me if so.

Pretend that F represents a pair of females and M represents a pair of males.

Consider a straight line arrangement. We can have either
M F M F
or
F M F M

There are 2 outcomes.

Now, effectively the 4 men have 4 different places they can go. (e.g. f f _ _ f f _ _ )
4! possible outcomes.

Similarly, so can the women. 4! possible outcomes.

Hence, in a straight line formation the answer is 2(4!)(4!) = 576


But we are talking about a circular arrangement, where there is no start. We need to cater for this by dividing by 4.
Hence 2(4!)(4!)/4 = 288

 

[trecex1]

I have a feeling I may be wrong. Someone correct me if so.

Pretend that F represents a pair of females and M represents a pair of males.

Consider a straight line arrangement. We can have either
M F M F
or
F M F M

There are 2 outcomes.

Now, effectively the 4 men have 4 different places they can go. (e.g. f f _ _ f f _ _ )
4! possible outcomes.

Similarly, so can the women. 4! possible outcomes.

Hence, in a straight line formation the answer is 2(4!)(4!) = 576


But we are talking about a circular arrangement, where there is no start. We need to cater for this by dividing by 4.
Hence 2(4!)(4!)/4 = 288

Could you explain why divided by 4? Also if it is a circle wouldn't your 2 outcomes be the same? They would not look different around a circle as opposed to a straight line. The provided solutions did: M1. 2x3!x4! = 288
M2. 4C2/2! x 2! x 4! = 288
M3. 3 x 2! x 2! x 4! = 288

 

[InteGrand]

Could you explain why divided by 4? Also if it is a circle wouldn't your 2 outcomes be the same? They would not look different around a circle as opposed to a straight line. The provided solutions did: M1. 2x3!x4! = 288
M2. 4C2/2! x 2! x 4! = 288
M3. 3 x 2! x 2! x 4! = 288

Call the males A, B, C, D. Note in the table's arrangement, it'll be:

--M
F---F
--M,

where M refers to a pair of males and F refers to a pair of females.

We can wlog (by rotational symmetry) place A at the top M above. We can choose this position in 2 ways (either A will have a male on his left, or on his right).

Now once A is placed, we have to place the remaining three males in the three seats left for males, which can be done in 3! ways.

Once all the males are placed, the females can go in their seats in 4! ways.

So the answer is 2*3!*4! = 288.

 

[leehuan]

I thought it was 2*3!*4! But as I went to post up my solution I just instantly forgot how I arrived at that conclusion...


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