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?
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.
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.
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.
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...