A bag contains 24 marbles, 4 red, 12 green, and 8 brown. How many ways can 5 marbles be drawn with all 5 marbles green.

I know you can consider that there are just 12 green marbles and 12 not green marbles but I am not sure of the steps to follow


You can see it like this way:

You know you have $12$ green marbles, $4$ red marbles and $8$ brown marbles, since you need to choose $5$ green marbles, you selected them from the $12$ possibles, so you need:

$$\binom{12}{5} = \frac{12!}{5!(12-5)!} = 792$$

Now that you have this done, you need to choose $0$ brown marbles from the $8$ possibles i.e.

$$\binom{8}{0} = 1$$

and $0$ red marbles from $4$ possibles, that is:

$$\binom{4}{0} = 1$$

OR you can see it as you need to choose $0$ marbles from the $12$ not green marbles possibilities ($8$ browns plus $4$ reds), i.e.

$$\binom{12}{0} = 1$$

Finally, all the possibilities to choose 5 green marbles are:

$$\binom{12}{5} \binom{8}{0} \binom{4}{0} = 792*1*1 = 792 = 792 * 1 = \binom{12}{5} \binom{12}{0}$$

So the answer is $792$ possible ways.


${12\choose 5}{8\choose 0}{4\choose 0}=\frac{12!}{7!\cdot5!}$.


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