I'm stuck in two ways of thinking

16! / (4!)^4

or

12! / (4! * 3!^3) * 5

Or I could just be completely wrong in my rationale.

Any help would be appreciated. I feel like I'm mixing something up or just forgetting something.

Thanks

- Thread starter Bob2424
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I'm stuck in two ways of thinking

16! / (4!)^4

or

12! / (4! * 3!^3) * 5

Or I could just be completely wrong in my rationale.

Any help would be appreciated. I feel like I'm mixing something up or just forgetting something.

Thanks

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Your thinking is on the right tracks. But I think that the question is ill posed.A class is going on a trip, there are 12 students and 4 teachers and 5 tents. Each tent will hold 3 students and 1 teacher how many groupings can be made?

16! / (4!)^4 or 12! / (4! * 3!^3) * 5

Not sure what the fifth tent has to do with anything. Does each tent have to have three students and one faculty?

Both my wife & I went to single sex secondary schools and colleges. We have two children, a daughter & son, who went to single sex boarding prep schools. So for this family this question is perfectly natural. But I don't think the statement is natural for today's students.

So lets say there are four tents which have three students and one faculty member.

There are \(\dfrac{12!}{(3!)^4}\) ways to assign twelve students to four faculty in groups of three.

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Isn't your result of \(\dfrac{12!}{(3!)^4}\) just the number of ways to arrange the 12 students??Your thinking is on the right tracks. But I think that the question is ill posed.

Not sure what the fifth tent has to do with anything. Does each tent have to have three students and one faculty?

Both my wife & I went to single sex secondary schools and colleges. We have two children, a daughter & son, who went to single sex boarding prep schools. So for this family this question is perfectly natural. But I don't think the statement is natural for today's students.

So lets say there are four tents which have three students and one faculty member.

There are \(\dfrac{12!}{(3!)^4}\) ways to assign twelve students to four faculty in groups of three.

Your thinking is on the right tracks. But I think that the question is ill posed.

Not sure what the fifth tent has to do with anything. Does each tent have to have three students and one faculty?

Both my wife & I went to single sex secondary schools and colleges. We have two children, a daughter & son, who went to single sex boarding prep schools. So for this family this question is perfectly natural. But I don't think the statement is natural for today's students.

So lets say there are four tents which have three students and one faculty member.

There are \(\dfrac{12!}{(3!)^4}\) ways to assign twelve students to four faculty in groups of three.

When I asked about the fifth tent having to be considered in finding the solution the response I got was each tent will have 3 students and one teacher, so I'm guessing the fifth tent is doing its job of just confusing my thought process...lol

My second thought on my original post was messed up with the 5th tent. I originally wanted to tackle the problem like so:

12! / (3!)^3

But I noticed that you have your denominator raised to the 4? can you explain?

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Well yes, that is the number of ways to divide twelve students into four groups of three each named \(A,~B,~C,~\&~D\) for teachers Adams, Burks, Clayton,& Dallas.Isn't your result of \(\dfrac{12!}{(3!)^4}\) just the number of ways to arrange the 12 students??

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Wait a minute. Imagine having the 4 groups of three students in a tent numbered 1-4. You can then place Adams in any of the 4 tents and then Burks into any of the three remaining tents, etc. Why wouldn't your answer be multiplied by 4!? If it has already been multiplied by 4! then can you please point it out? Thanks.Well yes, that is the number of ways to divide twelve students into four groups of three each named \(A,~B,~C,~\&~D\) for teachers Adams, Burks, Clayton,& Dallas.

If I'm understanding then it could be solved as C(12,3)*C(4,1)*C(9,3)*C(3,1)*C(6,3)*C(2,1) meaning it could be 12!4!/**(**3!3!3!**)**

However, I think this would result in over-counting by 4! because the tents are indistinguishable.

So it would be 12!4!/**(**3!3!3**!**4!**)** with the 4! canceling getting 12!/(3!)^3.

However, I think this would result in over-counting by 4! because the tents are indistinguishable.

So it would be 12!4!/

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OK, I get it. Thanks for your help.How many ways we arrange the string \(AAABBBCCCDDD~?\)

That is the number of ways to organize twelve people into four different cells, three to a cell.

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If we want to form four teams \(\mathcal{A,~B,~C,~D}\) of three students each from twelve students.If I'm understanding then it could be solved as C(12,3)*C(4,1)*C(9,3)*C(3,1)*C(6,3)*C(2,1) meaning it could be 12!4!/(3!3!3!)

However, I think this would result in over-counting by 4! because the tents are indistinguishable.

So it would be 12!4!/(3!3!3!4!)with the 4! canceling getting 12!/(3!)^3.

That can be done in \(\dfrac{12!}{(3!)^4}\).

If we

This difference is known as