# counting rule?

• August 5th 2010, 11:18 AM
sfspitfire23
counting 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?

• August 5th 2010, 11:21 AM
yeKciM
there's 10 combinations :D

$\displaystyle \frac {5!}{3!\cdot2!}=10$
• August 5th 2010, 11:21 AM
Unknown008
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.
• August 5th 2010, 11:31 AM
undefined
Quote:

Originally Posted by sfspitfire23
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?

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.
• August 5th 2010, 11:40 AM
yeKciM
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
• August 5th 2010, 11:45 AM
undefined
Quote:

Originally Posted by yeKciM
there's 10 combinations :D

$\displaystyle \frac {5!}{3!\cdot2!}=10$

Quote:

Originally Posted by yeKciM
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

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 $\binom{5}{2}$.

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.
• August 5th 2010, 11:45 AM
Unknown008
Quote:

Originally Posted by undefined
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.

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...
• August 5th 2010, 11:51 AM
yeKciM
Quote:

Originally Posted by undefined
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 $\binom{5}{2}$.

but that is also and $\binom{5}{3}$ so it's 5 shirts , 3 hangers , and 10 different combinations (ways) to hang shirts (2 who don't have any hanger don't get to hang, or that's what i'm missing here )
• August 5th 2010, 11:55 AM
undefined
Quote:

Originally Posted by yeKciM
but that is also and $\binom{5}{3}$ so it's 5 shirts , 3 hangers , and 10 different combinations (ways) to hang shirts (2 who don't have any hanger don't get to hang, or that's what i'm missing here )

I added an "edit" to the post that you quoted, which addresses this. Maybe you're familiar with the notation that you wrote C(5,3) while Unknown008 wrote P(5,3).
• August 5th 2010, 12:03 PM
yeKciM
Quote:

Originally Posted by undefined
I added an "edit" to the post that you quoted, which addresses this. Maybe you're familiar with the notation that you wrote C(5,3) while Unknown008 wrote P(5,3).

okay, :D but here (where i live) all problems that we get in that type of tasks is based on these that i wrote, and that's why I have assumed, and wrote that there :D Thanks :D for clarification