Permutation and combination difference, a flower shop question I have got a perm/combination question

In a certain flower shop, only 3 vases of flowers and 1 wreath can be displayed in the front window at a time. If there are 10 vases of flowers and 4 wreaths to choose from, how many different arrangements of vases and wreaths are possible? 

To my understanding it will be 
Since there is only 1 place for 4 wreaths so total possible combinations with other 3 places are 4
Now For flowers we have 3 places out of 10 diff flowers, it will be 10C3  = ${10 \choose 3}$ = 120
So total possible  arrangements = 4*120 = 480
However another version states that flowers we have 3 places out of 10 diff flowers should be 10P3 that is 720 
Which one is correct and why?

The difference has to do with the order in which you place the flowers into the three places. If you are lining them up in the window, then the order matters, so you are picking a flower for the first place (10 choices) picking a different flower for the second place (you now have 9 choices), etc...
Then you have $10*9*8 = 10P3 = \frac{10!}{7!}$ ways to arrange the flowers. 
Since the wreaths have only one place to go, order does not matter. In the case that you did not care about the order of the flowers, you are trying to pick a set of 3 flower vases from the vases available. In this case, you pick a flower first ( from 10), pick a flower second (from 9), etc... and then you need to remove the order in which you picked them. To remove the repeats, you determine the number of orders of 3 things and divide that out. Then you have
$$ \frac{10*9*8}{3*2*1} = 10 C 3 = \frac{10!}{3!7!} $$
ways to choose the flowers. 
