combinations and permutations

A bag contains 6 red marbles, 6 blue marbles, 10 white marbles and 7 yellow marbles. You are asked to draw 4 marbles from the bag without replacement. In how many ways can you draw two blue marbles? what formula do you use for questions like these

I am going to play with this question a little

If ALL the b***s are different. I mean, you have blue 1 blue 2 etc then

Number of ways of choosing 2 blue from 6 is 6C2 = 15

Number of ways of choosing 2 from the others 23C2 = 253

Total = 15*253 = 3795 ways

The number of ways that 4 marbles can be drawn with no restrictions is 29C4 = 23751

So the probablility of drawing 2 blue is    3795 / 23751 = 0.1598    (4 dp)

This is the way it must be done if you are looking at probablilities

BUT you are not concerned with probablilities and I think below is what the question may really be asking.

Lets see,

6 red marbles, 6 blue marbles, 10 white marbles and 7 yellow marbles

you could have 

So there are 6 different combinations that could include 2 blue b***s BUT they are NOT all equally likely to be drawn.  

We want to draw 2 blue ones out of 6 and out of the other 23, draw any 2 of them.

So.....the total number of ways to do this is :

C(6,2) * C(23,2)  = 3795 ways