There are 9 balls in a bag - 3 green, 3 blue, and 3 red.

Three kids reach in and grab 3 random balls each.

How many different arrangements can they end up with?

**
Clarification:
**
The balls are indistinguishable, and if one kid got RGR, its the same distribution as if he got GRR.

**
Image credit:
**
http://www.freeimages.com/

The answer is 55.

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If $N(a,b,c)$ is the number of ways the first kid can have $a$ matching balls, the second kid can have $b$ matching balls, and the third kid can have $c$ matching balls, then:

$N(1,1,1) = 1$

$N(1,2,2) = 6$

$N(2,1,2) = 6$

$N(2,2,1) = 6$

$N(2,2,2) = 12$

$N(3,2,2) = 6$

$N(2,3,2) = 6$

$N(2,2,3) = 6$

$N(3,3,3) = 6$

So, the total is $6*7+1+12=\boxed{55}$