At Gram and Gramp’s this weekend we played dominos, and the question came up of counting a set of dominos. In particular, Gramps wanted to know how many dominos there were in a set with double twelves.
After we were done playing, we lined them all up in the “domino triangle” with and Emery and I figured it out. The domino triangle is simply an arrangement of dominos with double zero in one corner, the zero-one next to it, followed by zero-two (etc.), and with the double one at the end, followed by the double two (etc.). Here is a partial rendering of the triangle for our double nine set:
+-+ +-+ +-+ |0| |0| |0| +-+ +-+ ... +-+ |0| |1| |9| +-+ +-+ +-+ +-+ +-+ |1| |1| +-+ ... +-+ |1| |9| +-+ +-+ ... +-+ |9| +-+ |9| +-+
To start, let equal the number of the largest double (twelve, if our case). We’re looking for , which is the number of dominos in a set with largest double . Simply counting the dominos in the triangle gives us a values for our function, but not the one we want:
n | f(n) -----+------- 0 | 1 6 | 28 9 | 55 12 | ??
Before dealing with , let denote the number of doubles in the set. A quick glance at the triangle will show that the formula for this function is
because there is one double for each
Now thinking back to the “domino triangle”, it’s easy to see that is both the width (base) and the height of the triangle. If you recall from planar geometry, the area of a triangle is half it’s base times height:
So a first guess at would be this:
But that isn’t correct. Thinking back to the domino triangle quickly shows why. If you were to take two copies of the triangle to make a rectangle (the reason the triangle area formula works), you’d see that you’d have to offset them by one domino in one direction, so the rectangle is actually by . So the right formula is this:
Simplifying by inlining the formula for yields:
This turns out to be the correct formula, and yields 91 when its argument is twelve. Thus a set of double twelve dominos will have 91 dominos.
What was particularly striking as Emery and I worked through the formulas, is that Emery had more trouble with the larger multiplication (which he hasn’t done in school) than following the function notation, doing the step-wise evaluation of the functions, and even the function nesting.