# Figuring out how many ways 5 marbles can be drawn (combination/permutation problem)

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.

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