Given:
N black balls
b1, b2, b3, ..., bn
N white balls
w1, w2, w3, ..., wn
N boxes
A random guys comes in, and puts 2 balls in each box, what is the probability of putting a white ball and a black ball in each box?
My effort:
Find all the possible ways to divide the 2N balls into group of 2.
Then find all the possible combinations for a black and a white ball, which is the Cartesian product. N^2
Then answer is:
But it seems I am wrong?
The original problem is a bit different, N guests take off their shoes, and some random guy scrambles them into new pairs, what is the probability of putting them in pairs where a left shoe is assigned to a right shoe.
Perhaps trying to think about this problem in a black-white problem was not a smart decision?
Any help is appreciated as it seems I have a mistake?
It is really dangerous to apply raw formulas in probability, that is what leaves the most room for mistakes. The best is imho to recall what the formulas do and somehow to reprove them every time.
So you have $2n\times (2n-1)$ options for the two balls to go in the first box, which has to be divided by $2$ if the order inside the box doesn't matter (thanks @2chromatic). Then $(2n-2)(2n-3)$ for the second box (again divided by 2), etc.
Therefore the total number of ways to put the balls in the boxes is $\dfrac{(2n)!}{2^n}$.
Now if you are to put only black/white couples, then you have $n\times n$ options for Box 1, $(n-1)\times (n-1)$ options for Box 2, etc. A total of $n!\times n!$ options.
The probability is therefore $$\dfrac{2^n(n!)^2}{(2n)!}$$ which happens to be equal to $$\dfrac{2^n }{{2n\choose n}}$$