Combination Question with Identical Pieces

[member 428835]

Homework Statement

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?

Homework Equations

Nothing comes to mind

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)?

 

[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.

 

[member 428835]

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

 

[member 428835]

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

 

[haruspex]

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

Those all look right to me.

 

[member 428835]

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

 

[haruspex]

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

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.

 

[member 428835]

sum over number of blue bricks (what bounds?), etc.

0 to 6?

 

[haruspex]

0 to 6?

You have already chosen r bricks to be red.

 

[member 428835]

You have already chosen r bricks to be red.

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.

 

[haruspex]

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.

Yes.

 

[member 428835]

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

 

[haruspex]

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

How do you get that from the nested sum?

 

[member 428835]

Nevermind, I made a dumb mistake. Thanks for your help.

 

[haruspex]

Nevermind, I made a dumb mistake. Thanks for your help.

You're welcome.