5 count bag of balloons, 6 color possibilities

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The discussion focuses on the distribution of colors in a 5-count bag of balloons that contains six color options. It establishes the probability of receiving only one color as 1 out of 1,296 bags and calculates the likelihood of getting all five different colors as 120 out of 1,296 bags. The main inquiry is about determining the probabilities for receiving 2, 3, and 4 colors in a single bag. There is speculation about whether Stirling numbers might be applicable for these calculations. The conversation highlights the complexities of color distribution in the balloon assortment.
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We have a standard assortment of balloons which is 6 different colors. We also have a 5 count bag of balloons which uses this standard assortment. Obviously you can get all 6 colors in a 5 count bag but my colleagues were curious about the other distributions.

The chances of only getting one color is pretty straight forward 1 out of 6^4 or 1 out of 1,296 bags will only have one color. Likewise getting 5 different colors is easy to figure out (1 * 5/6 *4/6 * 3/6 * 2/6) 120 out of 1,296 bags. My question to you is the other possibilities...2, 3, and 4 colors? Does this involve Stirling numbers?

1 color = 1/1,296
2 colors = ?
3 colors = ?
4 colors = ?
5 colors = 120/1,296

Thanks!
 
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cad08cad08 said:
... Obviously you can get all 6 colors in a 5 count bag

Hm ... maybe YOU can but I don't see how.
 
Helpful, thanks.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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