Statistics (probability) problem

[Lateral]

Homework Statement

Basically the garbage information summed up: a truck is transporting a bunch of batteries, with lots of different brands. The truck crashes and the batteries and scrambled.

Suppose the number of pairs of batteries is n. Each pair is a different brand. If 2r batteries are chosen at random, and 2r < n, show that the probability there is no matching pair, by brand, is:

2^{}2r n! (2n - 2r)! / (2n)! (n - 2r)!

Homework Equations

I don't know, but I can see that the expectation formula E(x) = \Sigma (x P(X=x)) and the Permutation formula nPr = n! / (n - r)! and Combination formula nCr = n! / (n - r)! r! may be useful as they seem to be of a similar format to the given equation.

The Attempt at a Solution

The part is struggle most with is why they have taken 2 to the power of 2r. (2n - 2r) is obviously total batteries less chosen batteries. The denominator is total batteries by pairs less batteries, but again I'm at a loss as to why they have done this.

Can anyone shed some light?

 

[EnumaElish]

2^(2r) = # of all subsets of a set with 2r elements.

n! / [(2r)! (n - 2r)!] = nC(2r)

(2n)! / [(2r)! (2n - 2r)!] = (2n)C(2r)

Does this help?