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In a department store at the mall, black and brown gloves are on sale. There are $N$ (identical) pairs of black gloves and $N$ (identical) pairs of brown gloves. If $N$ customers come in, one at a time and randomly choose and buy $2$ pairs each, find the probability of event $A$: each customer buys $2$ pairs of different colors (one black and one brown).

My attempt: $2N$ trials, $N$ choices out of $N$ black gloves, $N$ choices out of $N$ brown gloves, but I don't know what to do next. Any help is appreciated!

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Hint: What is the chance that the first customer buys different color pairs? Given that one pair of each color is gone, what is the chance the second customer buys different color pairs? Keep going.

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  • $\begingroup$ Thank you, it was a good hint! $\endgroup$ – AlphaDelphi Nov 12 '18 at 17:06
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When the first customer comes in and selects the first pair of gloves, nothing ye could have gone wrong from the goal of each customer selecting one brown pair and one black pair.

When the first customer selects the second pair there are $\frac {N} {2N - 1}$ pairs that are the opposite color from the pair she first selected.

Assuming the streak is still alive for customer two, that second customer can't break the streak with their first selection of a pair of gloves. But when the second customer selects his second pair, there are $\frac {N - 1} {2N - 3}$ pairs that are the opposite color of the first pair.

Continuing this pattern, and multiplying, you'll have:

$\frac {N \cdot (N - 1) \cdot (N - 2) \cdot ... \cdot 1} {(2N - 1) \cdot (2N - 3) \cdot (2N - 5) \cdot ... \cdot 1}$

Not sure how best to simplify that, maybe this? $N! \cdot \frac { N! 2^N} {(2N)!}$

The fraction on the right should cancel out all of the even factors in $(2N)!$

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  • $\begingroup$ Thank you for the help! $\endgroup$ – AlphaDelphi Nov 12 '18 at 17:07
  • $\begingroup$ @Kurt Schwanda, The fraction on the right don't cancel out all of the even factors in $(2N)!$ $\endgroup$ – Dhamnekar Winod Nov 29 '18 at 6:44
  • $\begingroup$ I may not be wording that well, and that could be the problem. (N! * 2^N) / (2N)! = ((N)*2 * (N-1)*2 * (N-2)*2 * (N-3)*2 * ... * (1)*2) / ((2N) * (2N - 1) * (2N - 2) * (2N - 3) * ... * 1) = ((2N) * (2N - 2) * (2N - 4) * (2N - 6) * ... * 2) / ((2N) * (2N - 1) * (2N - 2) * (2N - 3) * ... * 1) = 1 / ((2N - 1) * (2N - 3) * ... * 1) But I thought the numerator would cancel out the even factors from the denominator in (N! * 2^N) / (2N)!. Each factor of N! can be paired with a 2 from 2^N. $\endgroup$ – Kurt Schwanda Nov 29 '18 at 14:45

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