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The question : If i have a box that for example have 3 different color ball (R, G, B)

I would like to know how many combination of color that can occur in specific circumstances

ex :

If I take out the ball randomly and put it back, 5 times, how many combination of color can i have

(5R, 5G, 5B, 4R1G, 4R1B,....., 1R1G3B) *(4R1G=1G4R)

I could arrange it one by one but what if i have n color and I want to take "x" times,

Is there any Equation that i could plugin n (color) x (times being taken) And can it be derived, or if some one already ask something like this, could i get the link, i dont know what this topic of the problem called

Chrystian N
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  • In your example, there are three ways to select the color during each trial. If you select a ball five times, how many choices does that give you? – N. F. Taussig May 03 '18 at 13:20
  • Sorry I'm not native in English, I don't know what do you mean, if sample size i guess it's 3^5 ? i don't quite understand, please define "choices" a little bit – Chrystian N May 03 '18 at 13:29

2 Answers2

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You have $3^5$ possible arrangements (3 choices of color each time you pick a ball)

Vasili
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It appears to me that you are asking simply for the number of distinct outcomes, so that we only care about how many balls of each color are picked. In that case, the answer is given by stars and bars. In this problem the specific answer is $$\binom{5+3-1}{5} =\binom{7}{5}=\frac{7\cdot6}{2\cdot1}=21$$

For $n$ colors and $x$ picks the general answer is $$\binom{x+n-1}{x}$$

saulspatz
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