# combinations

• September 3rd 2008, 05:03 PM
kid funky fried
combinations
There are 18 people which are to be divided into 3 groups of six. How many ways can this be done?

This is a problem that is similar to a problem we had for h/w and quiz. I thought I understood when there were groups of two but now I clearly am at a loss. I thought it would start like:

$
{}_{18}C_6
$

How would you find the answer to this problem?
• September 3rd 2008, 06:36 PM
mr fantastic
Quote:

Originally Posted by kid funky fried
There are 18 people which are to be divided into 3 groups of six. How many ways can this be done?

This is a problem that is similar to a problem we had for h/w and quiz. I thought I understood when there were groups of two but now I clearly am at a loss. I thought it would start like:

$
{}_{18}C_6
$

How would you find the answer to this problem?

The answer is ${18 \choose 6} \, {12 \choose 6} \, {6 \choose 6}$. Do you see why?
• September 3rd 2008, 06:46 PM
kid funky fried
I think so...
First there was 18 people, so choose 6. Next, since we had already chosen 6 there would be 12 people left so it would be 12C6. Finally,
6 people remaining of which 6 were chosen.
• September 3rd 2008, 06:48 PM
mr fantastic
Quote:

Originally Posted by kid funky fried
First there was 18 people, so choose 6. Next, since we had already chosen 6 there would be 12 people left so it would be 12C6. Finally,
6 people remaining of which 6 were chosen.

(Yes)
• September 3rd 2008, 07:23 PM
kid funky fried
Thanks...
Thanks Mr. Fantastic, I appreciate your help.
I am glad to see I am on the right track.
• September 4th 2008, 03:49 AM
Plato
Quote:

Originally Posted by kid funky fried
There are 18 people which are to be divided into 3 groups of six. How many ways can this be done?

Quote:

Originally Posted by mr fantastic
The answer is ${18 \choose 6} \, {12 \choose 6} \, {6 \choose 6}$. Do you see why?

Actually this is an over count by a factor of (3!).
These are known as unordered partitions.
If the question were “how many ways to choose a red team, blue team, and green team?” then the above answer would be correct. That is because the partitions are ordered by color.

It is curious to note that the correct answer is equal to ${17 \choose 5} \, {11 \choose 5}$.
That is, take the first person listed and then choose five others to be with that person.
Then take the next person on the list not chosen and choose five more to go with that person.
Having done that, the remaining six people constitute the third group.
• September 4th 2008, 01:18 PM
kid funky fried
Plato,

I appreciate the response.
• September 4th 2008, 01:49 PM
mr fantastic
Quote:

Originally Posted by Plato
Actually this is an over count by a factor of (3!).
These are known as unordered partitions.
If the question were “how many ways to choose a red team, blue team, and green team?” then the above answer would be correct. That is because the partitions are ordered by color.

It is curious to note that the correct answer is equal to ${17 \choose 5} \, {11 \choose 5}$.
That is, take the first person listed and then choose five others to be with that person.
Then take the next person on the list not chosen and choose five more to go with that person.
Having done that, the remaining six people constitute the third group.

I wonder if the question is badly worded ......? Perhaps the OP can post the answer to the question once it becomes available.
• September 4th 2008, 02:10 PM
Plato
I do not consider this “There are 18 people which are to be divided into 3 groups of six. How many ways can this be done?” a badly worded question. In fact, in a course on combinatorics it is standard in the section discussing ordered verses unordered partitions of a set.
To partition a set of N into k groups (k|N=j) can be done in $\frac{{N!}}{{\left( {j!} \right)\left( {k!} \right)^j }}$ ways. That is an unordered partition
• September 4th 2008, 03:16 PM
mr fantastic
Quote:

Originally Posted by Plato
I do not consider this “There are 18 people which are to be divided into 3 groups of six. How many ways can this be done?” a badly worded question. In fact, in a course on combinatorics it is standard in the section discussing ordered verses unordered partitions of a set.
To partition a set of N into k groups (k|N=j) can be done in $\frac{{N!}}{{\left( {j!} \right)\left( {k!} \right)^j }}$ ways. That is an unordered partition

You're quite right of course. I meant badly worded in the sense that maybe the question didn't intend the idea of an unordered partition (it wouldn't be the first time I've come across questions that plainly say one thing but actually mean a different thing). I was just wondering ..... I appreciate you catching my error (one day I'll learn to stay away from these sorts of questions).
• September 4th 2008, 04:39 PM
kid funky fried
Quote:

Originally Posted by kid funky fried
There are 18 people which are to be divided into 3 groups of six. How many ways can this be done?

This is a problem that is similar to a problem we had for h/w and quiz. I thought I understood when there were groups of two but now I clearly am at a loss. I thought it would start like:

$
{}_{18}C_6
$

How would you find the answer to this problem?

The actual problem on the quiz is begins as:

2. (____) persons are to be grouped into three clubs, (____) in each club. In how many ways can this be done?

I omitted the numbers, as not all students have completed the quiz at this time.

The answer Mr. Fantastic gave was considered correct.

The reason for my confusion was because after posting the question on our school's board I had found that some had started the problem with 18C12 and I initially did not realize it was the same as 18C6.

The wording of the question I posted was slightly different. If this created a misunderstanding I apologize for the error and will strive for clarity in future posts.
• September 4th 2008, 10:26 PM
mr fantastic
Quote:

Originally Posted by kid funky fried
The actual problem on the quiz is begins as:

2. (____) persons are to be grouped into three clubs, (____) in each club. In how many ways can this be done?

I omitted the numbers, as not all students have completed the quiz at this time.

The answer Mr. Fantastic gave was considered correct.

The reason for my confusion was because after posting the question on our school's board I had found that some had started the problem with 18C12 and I initially did not realize it was the same as 18C6.

The wording of the question I posted was slightly different. If this created a misunderstanding I apologize for the error and will strive for clarity in future posts.

We'll forgive you this time (Rofl). Actually I think this thread is better in (terms of what can be learned from it) than it would have been had you not posted in the way you did.