Calculating the probability of a combination with repetition

Question

If I have n colors, (Let's say n=3 , blue=0, red=1 and green=2). And we have r boxes (Let's say r=5).

So we have these combinations (with repetition).

0 0 0 0 0  (all blue)
0 0 0 0 1  (1 is red, others blue)
0 0 0 0 2  (1 is green, others blue)
0 0 0 1 0  (1 is red, others blue)
...
...
2 2 2 2 0 (1 blue, others green)
2 2 2 2 1 (1 red, others green)
2 2 2 2 2 ( All green, this is the last combination)

If we want to count the possible permutations with repetition , P = n^r = 3^5 = 243

My question is , if we want to count the probability of an event where we have, say a greens, b red and c blue.

if we want to know for example the probability of these events : 02200, 02002, 22000, where we find 3 blues and 2 greens.

or the protability of having : two greens, 1 blue and 2 reds.

How can we calculate it regardless the order of the combination (where 02200 = 02002 = 22000).

Answer 1

The probability of picking one sequence that you're after is $1/3^5$ since each draw is independent. Then you just need to count how many other sequences match the critera of three zeros and two twos, say. For $n$ objects with $k$ types, and $n_i$ objects of type $i$ we have the number of possible permutations is $$\frac{n!}{n_1!\ldots n_k!},$$which in this case gives us $5!/(2!3!)=10$ thus the final probability is $$\left(\frac{1}{3}\right)^5\frac{5!}{2!3!} = \frac{10}{243} \approx 0.0412$$