There are 3 hangers and 5 shirts in a room. In how many ways, can these 5 shirts be arranged among the three hangers?

Would we solve this with the counting rule where we take 5(3)=15?

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- Aug 5th 2010, 11:18 AMsfspitfire23counting rule?
**There are 3 hangers and 5 shirts in a room. In how many ways, can these 5 shirts be arranged among the three hangers?**

Would we solve this with the counting rule where we take 5(3)=15?

- Aug 5th 2010, 11:21 AMyeKciM
there's 10 combinations :D

$\displaystyle \displaystyle \frac {5!}{3!\cdot2!}=10$ - Aug 5th 2010, 11:21 AMUnknown008
On the first hanger, you can put one from any of the five shirts. This gives 5 ways.

On the second hanger, you can put one from any of the four remaining shirts. This gives 4 more ways.

On the third hanger, you can put one from any of the three remaining shirts. This gives 3 more ways.

In total, you have 5x4x3 = 60 ways. - Aug 5th 2010, 11:31 AMundefined
I don't think it's clear what assumptions we're supposed to make. Supposing we can put multiple shirts on a hanger, a hanger can be empty, we must use all shirts, hangers are identical, and shirts are identical, there are 5 ways.

{0,0,5}

{0,1,4}

{0,2,3}

{1,1,3}

{1,2,2}

If the shirts are non-identical (which is a natural assumption), it becomes 41 ways

{0,0,5} --- 1

{0,1,4} --- 5

{0,2,3} --- C(5,2)

{1,1,3} --- C(5,3)

{1,2,2} --- 5*C(4,2)/2

and there are many variations of assumptions. - Aug 5th 2010, 11:40 AMyeKciM
hmmmmm...

how can it bee so many ways when there is only 10 ways to to put them on 3 hangers ... if position of shirt on hanger doesn't have any influence - Aug 5th 2010, 11:45 AMundefined
You wrote the number of ways to arrange five non-identical shirts on two identical hangers such that one hanger has three shirts and the other has two shirts. This is also just $\displaystyle \binom{5}{2}$.

Edit: Alternatively, you wrote the ways to choose three shirts from a set of five non-identical shirts and hang them on three identical hangers, one per hanger. Unknown008's variation is that the hangers are considered non-identical. - Aug 5th 2010, 11:45 AMUnknown008
Interesting... I didn't look at this question like this.

Anyway, I tackled this question like the questions I did in the past. The question mentioned 'arrangements' and by experience, I've learned it the hard way that the assumptions are: the shirts are non identical and the order of the hangers is relevant. And lastly, only three shirts get a place on the hangers.

Questions on combinations and permutations are not always explicit and can be interpreted in different ways... - Aug 5th 2010, 11:51 AMyeKciM
- Aug 5th 2010, 11:55 AMundefined
- Aug 5th 2010, 12:03 PMyeKciM