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:
How would you find the answer to this problem?
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 .
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 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 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).
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.