There are \(n\) (distinct) pairs of gloves. In how many ways can \(n\) people select a left handed and right handed glove, such that they do not select a pair? Please give me an expression in terms of \(n\).

There are \( n! \) ways for the people to first select a left-hand glove. Selecting a right hand glove is then equivalent to finding the number of derangements.

@Leonardo Cidrão
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No, that doesn't make sense. There aren't exactly \(n!\) ways for the people to select gloves such that at least one of them has a pair. Like Daniel commented, there are \(n!\) ways to select the left-hand gloves, and the number of ways to select the right-hand gloves is the number of derangements of \(n\) elements, which is denoted by \(!n\). It turns out, if you take \[\frac{n!}{e}\] (where \(e\) is Euler's number) and round it to the nearest integer, then you get \(!n\), for all positive integers \(n\).

Your question is incomplete. You are not mentioning whether each person takes equal no(2) of gloves or anyone can select any no. of gloves.
If there is no constraint on the no. of gloves a person can choose , then ans. is \( \binom{n}{2}^2 \)

I see that each person selects "a left-handed and a right-handed glove", indicating at least one of each? And by the condition that there are only \(n\) pairs of gloves and \(n\) people, it's pretty obvious that each person must select one of each. Of course, if the question was edited after you posted your comment then I can't tell.

Besides, even with your interpretation, for \(n=2\) it's wrong; there are \(2\) ways (a person chooses one left glove and the right glove that doesn't make a pair and the other person takes the rest; there are two such pairs) while your formula gives \(1\).

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TopNewestThere are \( n! \) ways for the people to first select a left-hand glove. Selecting a right hand glove is then equivalent to finding the number of derangements.

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But I think it's n! for lefthands and n! for the rights, in total of (n!)^2, isn' it?

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Except each person must have a left hand glove and a right hand glove that aren't a pair.

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Euler's number) and round it to the nearest integer, then you get \(!n\), for all positive integers \(n\).

No, that doesn't make sense. There aren't exactly \(n!\) ways for the people to select gloves such that at least one of them has a pair. Like Daniel commented, there are \(n!\) ways to select the left-hand gloves, and the number of ways to select the right-hand gloves is the number of derangements of \(n\) elements, which is denoted by \(!n\). It turns out, if you take \[\frac{n!}{e}\] (where \(e\) isLog in to reply

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Your question is incomplete. You are not mentioning whether each person takes equal no(2) of gloves or anyone can select any no. of gloves. If there is no constraint on the no. of gloves a person can choose , then ans. is \( \binom{n}{2}^2 \)

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I see that each person selects "a left-handed and a right-handed glove", indicating at least one of each? And by the condition that there are only \(n\) pairs of gloves and \(n\) people, it's pretty obvious that each person must select one of each. Of course, if the question was edited after you posted your comment then I can't tell.

Besides, even with your interpretation, for \(n=2\) it's wrong; there are \(2\) ways (a person chooses one left glove and the right glove that doesn't make a pair and the other person takes the rest; there are two such pairs) while your formula gives \(1\).

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