We don’t know for certain, but Jack suggests this:

William marries his maternal uncle’s widow, Hannah. Hannah’s brother (call him Fred) marries William’s widowed mother (call her Sue).

Husbands – William and Fred
Wives – Hannah and Sue
Brother/Sister – Hannah and Fred
Aunt/Nephew – William and Hannah
Mother/Father – Sue and Fred (if we say that Fred is the stepfather of William)

(Thanks, Jack.)

Solving this directly would require finding all the arrangements in which Sally and John are not side by side — 10 different cases, as it turns out.

But another way is to consider the opposite problem: What if Sally and John are required to stand side by side? That amounts to treating them as one child, though we must remember that there are two ways to “attach” them, Sally & John or John & Sally. Five children can be arranged in 5 × 4 × 3 × 2 × 1 = 5! = 120 ways, so there are 2 × 120 = 240 ways to arrange the six children so that Sally and John will be side by side.

Going back to the original problem, there would be 6 × 5 × 4 × 3 × 2 × 1 = 720 ways to arrange the six children if there were no constraints at all, so there must be 720 – 240 = 480 ways to line them up if Sally and John are to be kept apart.

(Shawn Godin, “Pólya’s Paragon: Au Contraire, Mon Ami,” Crux Mathematicorum 30:1 [February 2004], 13-15.)

02/09/2020 UPDATE: Remarkably, Marin Mersenne addressed the same question in his 1636 musical treatise Harmonie universelle. Reader Jon Wild explains that in a section on combinatorics, Mersenne asks how many ways there are of forming non-repeating melodies from the six notes of the hexachord ut-re-mi-fa-sol-la, then asks how many are “good,” which for him means they do not contain the interval (ascending or descending) of a major sixth, found between “ut” and “la.” He follows the same reasoning to the same answer, that 240 of the 720 melodies must be ruled out.

He actually writes out all 720 permutations, then explains that some of them are bad because they contain “ut la” or “la ut” (which is disagreeable, hard to sing, and painful to the ear, and whose frequency ratios are objectionable on numerological grounds). 120 of the listed permutations include the major sixth ascending and 120 descending; subtracting these from the 720 listed permutations leaves 480 good melodies.

(Marin Mersenne, Harmonie universelle, 1636, Proposition 9 from the “Deuxième livre des chants” [second book on Melody].) (Thanks, Jon.)

1. Qh3! and the queen will mate.

From Novosty, 1963.

If Alice is speaking the truth, then Bob and Charlie are liars. But in that case Bob would be uttering a true statement, which is a contradiction. So Alice is a knave. If that’s the case, then either Bob or Charlie (or both) is a knight. If Bob is a knave, then he’s lying when he says that there’s exactly one truth-teller among the other two. But that’s another contradiction: In that circumstance his statement would be right, since Alice is a knave and Charlie would be a knight. So Bob must be a knight, which means that we can trust what he says and Charlie is a knight too. So Charlie says, “One.”

(Khovanova’s students pointed out that the very question in the puzzle implies that Charlie makes a predictable answer, which wouldn’t be the case if he were a liar. The result follows from there.)

Obligatory John Finnemore sketch:

With 1. Ne4 White threatens to mate with 2. Rb8. Black can avoid this by capturing the b4 knight, and he has four ways to do this, but each blocks the b4 flight square, permitting mate:

1. … axb4 2. Rc5#
2. … Rxb4 2. Nc3#
3. … Bxb4 2. Bc4#
4. … Nxb4 2. Nd6#

Kenneth Schilling of the University of Michigan offered this solution: Call any set of citizens all of whom know one another a clique. Now if there’s a clique B that has fewer than n + 1 citizens, then by hypothesis some citizen who’s not in B knows everyone in B, and we can add that person to B to form a larger clique. That means that a clique exists of size n + 1. The set A of citizens who aren’t in B has size n. So by hypothesis there’s some person in B who knows every person in A, and hence that person knows everyone in the town.

(Ashay Burungale and Kenneth Schilling, “Getting to Know You, Directly and Indirectly,” American Mathematical Monthly 115:9 [November 2008], 862.)

All three players raised their hands, so each can see at least one red hat. This means that at least two of the hats are red; if two or more were blue then there’d be at least one player who didn’t raise her hand. But any player who can see a blue hat can immediately infer that her own hat must be red, because she can see a red-wearing player whose hand is raised. In the puzzle this doesn’t happen: All three players raise their hands and yet none of them makes this inference. That allows one of the players to conclude that none of them is wearing a blue hat; all three hats must be red.