$6$ distinguishable balls into $4$ indistinguishable boxes I got a problem that goes as follows:
With $6$ balls and $4$ boxes how many ways can we place the balls in the boxes if the balls are distinguishable and the boxes are not.
In this answer it was proposed (if I interpret it correctly) that the number of ways to do just that would be 
$$A=\sum_{r=0}^{4}S(6,r)$$
where $S(n,k)$ would be The Stirling numbers of the second kind.
I evaluated $A$ (according to table at the wikipedia page) to $187$, while for the answer of the entire question my teacher proposes $342$. What mistake have I made here?
 A: There are 8 possible distributions : (6 0 0 0) (5 1 0 0) (4 2 0 0) (4 1 1 0) (3 3 0 0) (3 2 1 0) (3 1 1 1) (2 2 2 0) (2 2 1 1).
(6 0 0 0) - only 1 combination.  
(5 1 0 0) - 6 ways of selecting the lone ball.  
(4 2 0 0) - 6x5/2 = 15 ways of selecting the pair.
(4 1 1 0) - again 15 ways of selecting the two lone balls.  
(3 3 0 0) - 6x5x4/3x2 = 20 ways of selecting a triple, but the two triples are indistinguishable so there are only 20/2 = 10 distinct combinations.
(3 2 1 0) - 6 ways of selecting the lone ball; 5x4/2 = 10 ways of selecting the pair; so a total of 6x10 = 60 combinations.
(3 1 1 1) - 6x5x4/3x2 = 20 ways of selecting the 3 indistinguishable lone balls.  
(2 2 2 0) - 6x5/2 x 4x3/2 = 90 ways of pairing the balls; the 3 pairs are indistinguishable so 90/3x2 = 15 combinations.
(2 2 1 1) - 6x5/2 = 15 ways of selecting the lone balls; 4x3/2 = 6 ways of selecting a  pair; the pairs are indistinguishalbe so 6/2 = 3 ways; in total 15x3 = 45 combinations.
Total = 1+6+15+15+10+60+20+15+45 = 187.
