Since there’s only one way to sum two numbers and get zero, this works out so that the total number of configurations is just the sum of the integers from 1 to N+1. Which is famously (N+1)(N+2)/2.

]]>Conjecture: if you have K boxes totaling N, the number of ways to construct the sum will be choose(N+K-1,K-1). I’m not sure how to prove that, but it’s consistent with the results for one and two boxes, as well as what I have tried for three boxes. Possibly a proof by induction will work, i.e., if it works for a particular choice of N and K, then it must work for N+1 and K, and N and K+1.

And that sounds like a great game to play with SteelyKid. Encourage her inner mathematician while you can.

]]>*‘Cause I love the players, and You! Love! Croquet!*

I’m not connected well with current popular music, so I have no idea what song produced that result, but I agree that it’s a brilliant mondegreen. Alas, it’s harder to do that when you are an adult listening to a song in your native language, but for an illustration of what Latin can do, look on Youtube for various versions of Carl Orff’s epic “Oh Four Tuna” (which I’m not linking to because most of them are NSFW).

]]>