So, we want to have 6 candies in a box compounded of 5 different sorts.



Now, the most simple reading/interpretation is that the order of these 6 candies 
in the box does not matter.
At least, it is so from the point of view of any child :)


If so, then we should have all 5 candies in the box of different sorts,
PLUS the 6-th candy of some sort.


So, for us the only question does matter: which of 5 possible sorts of candies is doubled?
It makes clear that the answer to the problem's question is 5  (in this problem's reading).


I am 97% sure that it is the correct reading (and, correspondingly, a correct answer), 
and it is the reading and the answer which is expected in this problem.


But formally, there is a small percentage opportunity for another reading and another
solution, which follows below.


This another reading says that there are 5 different letters A, B, C, D and E, that correspond 
to 5 sorts of candies, and the question is how many possible arrangements (6-letter words) 
can be written in this 5-letter alphabet so that all 5 letters are used.


So, in this problem there are 6! = 720 permutations, in all, but we should consider only 720/2 = 360 
of them as distinguished arrangements, since each permutation has one and only one letter doubled,
and the doubled letter produces only one arrangement from two possible permutations.

It is the second possible (3 percent of possibility) interpretation, the solution and the answer.


So, in first interpretation the answer is :  there are 5 different candy boxes (pointing only coinciding sorts of candies).

In the second interpretation, the answer is :  there are 360 distinguishable boxes/arrangements.


But not 720, as in the solution by @Theo, which is incorrect.