Arrange 6 fruits from three identical watermelons, three identical bananas and three identical strawberries. 
The Question


Marcus stumbles upon 3 identical watermelons, 3 identical bananas and 3 identical
strawberries. Marcus wishes to create an arrangement in a line using precisely 6 of these fruits. How many ways can he do this? (BWW is different to WBW).


My Attempt

We can choose 0 watermelons, 1 watermelon... 3 watermelons, so 4 possible ways.
Similarly, 4 ways to choose a number of bananas.
Similarly, 4 ways to choose a number of strawberries.
So 64 ways. But we must then remove combinations such as 0 watermelons, 0 bananas and 0 strawberries which have less than 6 fruits in the arrangement. So we subtract 25 (I counted).
Answer = 39?

Post-Note

I'm not sure if I'm right, I'm not sure if I accounted for the permutations correctly, and I definitely don't think this is the best way. Help appreciated!

Thanks!

 A: One way to do this is to sort by how many fruits are repeated. We get different answers depending on if fruits of the same type are distinguishable or not. From OP's answer it looks like watermelons are interchangeable with each other, bananas are interchangeable with each other, strawberries are interchangeable with each other, and so I will answer according to that logic.
Since there are 6 fruits in the line, there are 3 types of arrangement:

*

*One fruit does not appear: If one fruit doesn't show up at all, we have to use all 6 of the other two fruits. So there are 3 ways to choose the absent fruit, and $\binom{6}{3} = 20$ ways to pick the order of the other 2 fruits, for a total of $3 * 20 = 60$ arrangements.


*One fruit shows up once, one fruit shows up twice, one fruit shows up three times: I have $6$ choices for where to place the lone fruit, and then $\binom{5}{2} = 10$ choices left for where to place the fruit repeated twice. I have $3$ choices for the lone fruit and $2$ choices left for the fruit repeated twice, giving me $6 * 10 * 3 * 2 = 360$ such arrangements.


*Every fruit shows up twice: This is the number of different "words" possible from the "word" WWBBSS, which is $\frac{6!}{2!2!2!} = 90$.
So the correct answer, assuming any two watermelons are interchangeable, any two bananas are interchangeable, and any two strawberries are interchangeable, is $60 + 360 + 90 = 510$.
