Arrangements help (1 Viewer)

[darkphoenix]

Recently doing all sorts of 3and 4 unit papers, but only getting around 50 % on the arrangements and probability part. For example In how many ways can 4 men and 4 women be arranged around a circular table if all the men are in pairs separated by two pairs of women. How do you start this... please show each step and I really need tips to do these type of question. Thanks for help

 

[lolcakes52]

In general, exposure to lots of questions will really help with this.

For the question you posted, approach it like this.

How many ways can I make two pairs of 4 men/women?

Easy enough, its (4C2)*(2C2) per gender or (4C2)^2 in total.

So we have 4 groups of 2 like gendered people. How many ways can they be arranged amongst themselves ie in there pair?

Easy question, its 2 per pair or 2^4 in total.

Now we have to position the pairs around the table, so that the pairs of men will be separate.

First we pick one of the pairs, there are 2 ways of doing this. Then we pop them down on the table, there is only 1 way of doing this due to symmetry.

Now we plop down the two pairs of women, there are only 2 ways to do this. We finally place the last pair of men.

So multiplying all this together we get

(4C2)^2 * 2^4 * 2 * 2

I hope that is right. Ask away!

 

[asianese]

You can also think of placing the pairs as this:

Place 1 pair of men down. Now the two pairs of women must go either side of them. So there are 2 possibilities. But the pairs of men can swap positions to - 2 possibilities. This gets the same answer.

 

[RealiseNothing]

You can also think of placing the pairs as this:

Place 1 pair of men down. Now the two pairs of women must go either side of them. So there are 2 possibilities. But the pairs of men can swap positions to - 2 possibilities. This gets the same answer.

Doesn't the symmetry make this redundant, ie you double count?

 

[lolcakes52]

Doesn't the symmetry make this redundant, ie you double count?

It's basically the same thing that I did but phrased differently. If you want to approach it typically it is just 2C1 ways of choosing a group of men to place down. Then its 2C1 to choose the women to the left, 1C1 way to choose the next pair of men and finally 1C1 way to choose the group of women.

 

[darkphoenix]

Ummm, the answer is 288

 

[Shadowless]

3! x 4! x 2

 

[RealiseNothing]

It's basically the same thing that I did but phrased differently. If you want to approach it typically it is just 2C1 ways of choosing a group of men to place down. Then its 2C1 to choose the women to the left, 1C1 way to choose the next pair of men and finally 1C1 way to choose the group of women.

But the symmetry of the circle makes placing the men redundant. So to arrange the pairs it is just 2C1, 1C1, 1C1, 1C1. There isn't two 2C1's.

 

[darkphoenix]

don't get them.... sigh