# Combinatorics - box of white and black screws

1. Jan 2, 2012

### toofle

1. The problem statement, all variables and given/known data
The screwengineer Pelle has a box with 16 black screws and 16 white screws.
a. In how many ways can he pick an even(at least 2) amount of screws from the box?
b. Pelle randomly chooses one of the alternatives in (a), what's the chance that gets as many white as black screws?
It is allowed to answer in binomial coefficients.

2. Relevant equations

3. The attempt at a solution
a) easy, sum k from 2 to 32, 32 nCr k => 4294967263 ways.
b) this is the hard part. I have the solution but I don't understand it.
The nbr of nomempty sets with as many black as white screws are (16 nCr 1)^2+(16 nCr 2)^2+...+(16 nCr 16)^2
Divide that by the answer in (a) to get the answer to (b) => 0.28 = 28%
I don't get the part with nonempty sets.

2. Jan 2, 2012

### vela

Staff Emeritus
That's not correct. Why would you include the 32C3, for example, the number of ways to choose 3 screws?

Say you wanted to calculate the number of ways you could choose 1 white screw and 1 black screw. How would you do that?

3. Jan 3, 2012

### toofle

I meant sum from 2 to 32 with step=2, ie 2,4,6,8,...,32. for got to write that but the answer was calculated that way.

ofc, (16 nCr 1) * (16 nCr 1). then the same for 2,3,4 etc. I get it.

4. Jan 3, 2012

### vela

Staff Emeritus
Actually, you didn't calculate the answer that way, which is why I brought it up. The total number of combinations of choosing from 32 screws is 232=4294967296. If you want only the combinations with an even number of screws, you should get about half that or about 2147483648.

5. Jan 3, 2012

### toofle

actually you are right. I forgot to step with 2 in my TI82 as well.
Thanksm but I get it now.

Last edited: Jan 3, 2012