# Math Help - counting rule?

1. ## 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?

2. there's 10 combinations

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

3. 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.

4. 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.

5. 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

6. Originally Posted by yeKciM
there's 10 combinations

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

7. 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...

8. 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 )

9. 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).

10. 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, 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 Thanks for clarification