Permutations with 3 balls of different colours in 4 glass cylinders

[zorro]

Homework Statement

In how many ways can 3 balls of different colours be put in 4 glass cylinders of equal width such that any glass cylinder may have either 0,1,2 or 3 balls?

The Attempt at a Solution

Using the formula for no. of ways for distribution of n distinct things into r different groups when empty groups are allowed-

^n+r-1C_r-1 = ⁶C₂ = 15
Now the balls can be arranged among themselves in 3! ways
there fore 15 x 3! =90

The answer is wrong.
Help

 

[Tedjn]

Let the three balls be R, G, B. Let | denote a partition between glasses. Without the factor of 3!, you count R|GB|| but miss G|RB||. With the factor of 3!, you count R|GB|| and R|BG|| both when they are indistinguishable.

 

[jbunniii]

Let the three balls be R, G, B. Let | denote a partition between glasses. Without the factor of 3!, you count R|GB|| but miss G|RB||. With the factor of 3!, you count R|GB|| and R|BG|| both when they are indistinguishable.

Interesting. I interpreted the question to mean that R|GB|| and R|BG|| are distinguishable. I assumed that point of describing "glass cylinders of equal width" instead of "buckets" or "urns" was to indicate that if more than one ball was in a cylinder, they would be stacked vertically, so the order in which they were inserted into the cylinder matters. (Certainly one can visually distinguish between blue stacked on top of green versus the opposite.)

OP: which interpretation is correct?

 

[zorro]

OP: which interpretation is correct?

The question mentions about equal width of cylinders so that there is no biasing in the filling of cylinders (no where does it mention that the balls are stacked vertically lol).

I just figured that I did not use the formula correctly-
n = 3 and r = 4
^n+r-1C_r-1 = ⁶C₃ multiplied by 3! = 120 (which is correct)