# Combination Question with Identical Pieces

1. Apr 27, 2016

### joshmccraney

1. The problem statement, all variables and given/known data
You have 4 red, 7 blue, and 12 white bricks and grab a random handful of 6 bricks.
a) How many handfuls will contain exactly 3 red bricks?

b) How many handfuls will contain exactly 2 of each color?

c) How many handfuls will be all the same color?

d) How many handfuls will contain at least one red brick?
2. Relevant equations
Nothing comes to mind

3. The attempt at a solution
I assume bricks of the same color are identical.
a) {b b b}, {b b r,} {b r r},{ r r r} so I think there are 4 ways to get a hand of 3 red bricks. Is there a formula I should use?

b) Only one way to do this.

c) Only two ways to do this since there is not enough red.

d) I think at least one red is equal to the total number of handfuls minus handfuls of no reds. No reds implies only blue and white only. Can anyone help me on the setup here? I'm guessing a similar approach to the method in a), but perhaps there's a way to extrapolate this idea into an equation (combinations maybe)?

Could you tell me if you agree with a), b), and c)?

2. Apr 27, 2016

### haruspex

Not sure how to interpret "how many handfuls". I would think it is meant in a probabilistic sense, as in, in a very large number of trials how many? So I would treat all the bricks as fundamentally distinct.
If I had written the question in terms of handfuls (unlikely) and intended the approach you have used, I would have asked for "different-looking handfuls", or somesuch.

3. Apr 27, 2016

### joshmccraney

So then the answer to a) would be $4C3 \cdot 19C3$? If so I think I can redo all of them no problem.

4. Apr 27, 2016

### joshmccraney

So would we have
a) $4C3\cdot19C3$
b) $4C2\cdot7C2\cdot12C2$
c) $7C6+12C6$
d)$23C6-19C6$

5. Apr 27, 2016

### haruspex

Those all look right to me.

6. Apr 27, 2016

### joshmccraney

Cool, I thought so but wasn't sure if the blocks were identical how I'd do part d.

7. Apr 27, 2016

### haruspex

If you were just counting the different patterns, the method of subtracting the count with no red bricks from the unrestricted count does not help because you also have the restriction that there are no more than four red bricks. So you might as well do it is a nested sum. It is analogous to a double integral. Sum over number r of red bricks, 1 to 4, {sum over number of blue bricks (what bounds?), etc.}

Edit: added braces to nested sum for clarity.

8. Apr 27, 2016

### joshmccraney

0 to 6?

9. Apr 27, 2016

### haruspex

You have already chosen r bricks to be red.

10. Apr 27, 2016

### joshmccraney

Didn't see this, thanks. So $b \in [0,6-r]$ and $w \in [0,6-r-b]$ where my notation is a little lax.

11. Apr 27, 2016

### haruspex

Yes.

12. Apr 27, 2016

### joshmccraney

Awesome, thanks. So the answer is 36-7 = 29? Doesn't feel right.

13. Apr 27, 2016

### haruspex

How do you get that from the nested sum?

14. Apr 27, 2016