Permutations and combnations: groups of equal size (1 Viewer)

[Kurosaki]

Hi BOS denizens, I'd just like to ask, why is it then when one divides a group of people into x smaller groups of equal size, they divide the total number of ways by x factorial?

E.g. dividing 12 people into 3 groups of 4 gives

Why does one divide by 6 in this case? My teacher just says that the groups are 'interchangeable' but I don't really get what this means . Could someone please offer an intuitive explanation?

 

[Squar3root]

you have to divide by 6 because the people can be in either group so one person can be in one group when the next time he could be in the other group with different people if that makes any sense

 

[braintic]

Think of the groups as being labelled A, B and C.

12C4 is the number of ways of picking 4 people to go in group A (say). Then 8C4 is the number of ways of picking 4 more for group B. Then group C picks itself.

Now think of the people as being labelled a, b, c, d, e, f, g, h, i, j, k & l.
One possible outcome is (a,b,c,d) in group A, (e,f,g,h) in group B, and (i,j,k,l) in group C.
Another possibility is (e,f,g,h) in group A, (i,j,k,l) in group B, and (a,b,c,d) in group C.
But these two possibilities provide EXACTLY the same outcome (as the labelling of groups was artificial).
In fact there are 6 such identical outcomes ... the number of ways of arranging the letters A, B, C .... 3 factorial in other words.
So we divide by 6 because we are overcounting by a factor of 6.

If we were dividing 21 people into two groups of 3 and three groups of 5:
Firstly multiply 21C3 by 18C3 by 15C5 by 10C5 (by 5C5=1)
Then to account for overcounting, we divide by (2! times 3!)

 

[Kurosaki]

Think of the groups as being labelled A, B and C.

12C4 is the number of ways of picking 4 people to go in group A (say). Then 8C4 is the number of ways of picking 4 more for group B. Then group C picks itself.

Now think of the people as being labelled a, b, c, d, e, f, g, h, i, j, k & l.
One possible outcome is (a,b,c,d) in group A, (e,f,g,h) in group B, and (i,j,k,l) in group C.
Another possibility is (e,f,g,h) in group A, (i,j,k,l) in group B, and (a,b,c,d) in group C.
But these two possibilities provide EXACTLY the same outcome (as the labelling of groups was artificial).
In fact there are 6 such identical outcomes ... the number of ways of arranging the letters A, B, C .... 3 factorial in other words.
So we divide by 6 because we are overcounting by a factor of 6.

If we were dividing 21 people into two groups of 3 and three groups of 5:
Firstly multiply 21C3 by 18C3 by 15C5 by 10C5 (by 5C5=1)
Then to account for overcounting, we divide by (2! times 3!)

I see, that makes quite a bit more sense now, thanks .
If I could trouble you with another inquiry: I was doing a past paper question, and was wondering, why is it that they didn't divide 420 and 1680 by 3!=6 each? I know that it doesn't affect the calculation, but since you have three groups with the same people, shouldn't you divide 420 and 1680 by 6 each to get the correct number of ways?

This was my working: Choose one group from the three for the sisters to be in, which can be done in 3 ways. Choose someone to fill in the last place- 7 ways. get the rest of the groups- . Multiplying and dividing by 3! (since there are three groups with the same number of members), then we should get 70.
And for the total number of ways, it's simply right?

 

[braintic]

I see, that makes quite a bit more sense now, thanks .
If I could trouble you with another inquiry: I was doing a past paper question, and was wondering, why is it that they didn't divide 420 and 1680 by 3!=6 each? I know that it doesn't affect the calculation, but since you have three groups with the same people, shouldn't you divide 420 and 1680 by 6 each to get the correct number of ways?

This was my working: Choose one group from the three for the sisters to be in, which can be done in 3 ways. Choose someone to fill in the last place- 7 ways. get the rest of the groups- . Multiplying and dividing by 3! (since there are three groups with the same number of members), then we should get 70.
And for the total number of ways, it's simply right?

This can be done MUCH more simply, without resorting to perms and combs.

Place one of the sisters in a group. There are then 8 places left to fill, 2 of which are in the same group as the sister.
So the probability is 2/8 = 1/4

 

[seanieg89]

Yeah, they are glossing over that fact. In such situations it is usually clear when a factor will appear in both things you are counting (so you can ignore it), but for a worked solution that is a little sloppy. Their calculations are correct if you treat the three groups are ordered, eg you are choosing three groups of people in your team to do different jobs.



Note: For that question, you don't need to go through the work of counting all possibilities.

From the perspective of the first sister, she has 2 out of the 8 other people randomly selected to be in her group. The probability of the first one not being her sister is 7/8, and the probability of the second one also not being her sister is 6/7. So the probability of her sister being in her group is 1-(6/7)x(7/8)=1/4.

Edit. braintic says this in the above post in a cleaner way.

 

[braintic]

Edit. braintic says this in the above post in a cleaner way.

I notice that lots of people here talk about older posts being above.
But I always see the most recent post at the top, and the older posts below.
This is not a complaint - I prefer the latter view. But I am wondering why I am getting a different view to everyone else?

 

[seanieg89]

I notice that lots of people here talk about older posts being above.
But I always see the most recent post at the top, and the older posts below.
This is not a complaint - I prefer the latter view. But I am wondering why I am getting a different view to everyone else?

Interesting, I will have a look later as that setting would probably be ideal for me too. I am guessing it will be an option in the settings page (check out the link that is hopefully in the top right hand corner of any page on BoS, right next to log out).