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Suppose you have k different pairs of gloves (k left gloves and k right gloves, for 2k gloves in total) in your chest. You take the gloves out of the chest one by one without looking and lay them out in a row on the floor. What is the probability that no two matching gloves are next to each other?

I tried approaching this as follows:

There are 2k possible choices for the first glove, and then the next glove has 2k - 2 choices since matching pair can't be next to it, then the third has 2k - 2 choices as well because the 1st glove's counterpart can be used now. Now, for the 4th glove, there are 2k - 4 choices, and so forth, but I have a feeling this is incorrect. Any solution/pointer would be appreciated.

stephanie
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  • The number of choices you have for the fourth glove depends on whether or not the first and third gloves form a pair – Moko19 Jul 25 '21 at 06:26
  • You should apply principle of inclusion exclusion. For $2n$ gloves ($n$ pairs), you can write number of favorable arrangements as $ \displaystyle \sum \limits_{i=0}^n (-1)^i \cdot {n \choose i} \cdot 2^i \cdot (2n-i)!$ – Math Lover Jul 25 '21 at 08:43
  • One more https://math.stackexchange.com/questions/463035/no-husband-can-sit-next-to-his-wife-in-this-probability-question – Math Lover Jul 25 '21 at 08:45

2 Answers2

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We can more easily compute using the complement, viz. by counting all arrangements minus arrangements where matching pairs are adjacent.

The latter can be obtained by first "gluing together matching pairs" and treat each such pair as a unit, and then apply inclusion-exclusion

For illustration, suppose there are $4$ pairs of socks.

Total arrangements are $8!$ and arrangements with one pair chosen in $\binom 41$ ways, glued together (in $2$ possible ways, LR or RL), there would be $7$ units, and so on and so forth, yielding

$\dfrac{8! - \binom41 2^1 7!+2^2\binom42 6! - 2^3\binom435! +2^4\binom 44 4!}{8!}$

Addendum

This actually amounts to the relaxed menag$\dot{e}$ problem for seating married couples in a row so that no husband sits next to his wife. You can look it up for further insight.

true blue anil
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HINT: Start with small specific examples for $k$ and try to build a rule (and prove the rule holds for other examples)

If $k=2$, there are $24$ ways we can distribute the gloves, but only $8$ sequences where none of the pairs are adjacent (four choices for the first glove and two for the second). Overall probability is $\frac 13$

If $k=3$, there are $720$ ways we can distribute the gloves. We have six choices for the first glove, and four for the second. We then need to divide into two cases. If the third glove matches the first (which is possible in only one such way), we have two options for the fourth glove. If the third glove doesn't match the first (which is possible in two ways), we have two options for the fourth glove and two for the fifth. In total, that leaves $240$ total distributions, for an overall probability of $\frac{1}{3}$

If $k=4$, we have $8!$ total distributions. We have eight choices for the first glove and six for the second. Now, we divide into four cases. If the third glove matches the first and the fourth matches the second (which can happen only one way), we are left with two pairs and four spaces, which we stated above had $8$ sequences. If the third glove matches the first but the fourth doesn't match the second (which can happen four ways), it is analogous (do you see how) to the state with six gloves after you have placed the first two, which had $10$ sequences. If the third glove doesn't match the first and the fourth glove matches the first or the second, with (with a choice of four gloves for the third and two for the fourth), there are four remaining gloves, of which we have a single pair and two unpaired; we can arrange these in twelve ways (six ways to distribute the pair and two ways to distribute the unpaired gloves). In the final case, we have the first four gloves being one from each pair (with four choices for the third glove and two for the second), we then have three choices for the fifth glove, three for the sixth, and two for the seventh. In total, we have $\frac{8\cdot6(8+4\cdot10+8\cdot12+8\cdot18}{8!}$

Can you proceed from here?

Moko19
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