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Math Puzzle: Defective Question?

March 11, 2023 pm31 9:54 pm

McRib doesn’t do counting, so when a student brought him this tricky counting question, he asked me. And I thought his wording was ambiguous, so I asked him to get me the actual question. Now, a few weeks later Math Counts released the question, and he sent it to me. Take a look:

You might want to play with it before you read on.

I think there are two issues. One of them we might overlook – I often complain about it, and people usually tell me that we know what they meant. The second, though, looks to me like the author/editor had not sufficiently thought through their question.

See what you think.


There is an implied event here (George starts drawing segments until each point is doubly connected), but they don’t bother saying that. They don’t say that each “outcome” is equally likely. They don’t say that each possible segment is equally likely. I guess we could assume the former (and whether the latter is true or not we don’t really need to know) – but why not state it? Further, they really want to know how many convex hexagons George might get out of how many total outcomes George might get. They wrote “probability” but they really meant it in the A out of B sense. They just wanted a fraction.

What’s the event?

“If George draws line segments connecting pairs of points so that each point is connected to two other points” describes an event with many (how many?) possible outcomes –

“the resulting figure” describes a subset of those outcomes.

“a convex hexagon” describes a subset of those outcomes.

What did they miss?

When George follows their directions, he can create a loop, cycling through the 6 points. But he can also create two triangles.

But “the resulting figure” seems to exclude the cases with two triangles.

Work towards solution(s)

(I have put simpler work at the bottom – with nicer explanations)

Label the points U, V, W, X, Y, Z. Starting from U, George can connect to 5 points, then 4, then 3, then 2, then 1. 5×4×3×2×1= 5! But for every route, eg U-W-Z-V-X-Y there is its reverse U-Y-X-V-Z-W. So 5! counts each loop twice – there are only half that number of loops: 5!/2, or 60.

But nothing in the initial instructions stops George from connecting U to V, V to W, and W back to U. Kind of cool here, once he’s done, we know what his next three steps will be (X to Y, Y to Z, Z to X) – we don’t even have to map them out. So how many triangles can George make starting with U? He’s got 5 more points. He needs to take 2 of them. And {}_ 5C_{2} = 10.


Since they ask about the “resulting figure” (singular) it would seem they overlooked the triangles. So the answer might have been 1 good hexagon out of 60, or 2 good hexagons (one forwards one backwards) out of 120.

But if they meant for the triangles to be included, things get messier. I could claim the denominator should be 60 + 10 = 70, but there is more here than that.

Will we count ∆UVW and ∆XYZ as distinct from ∆UVW and ∆YZX? We can arrange all these letter twelve ways, fixing the U first. Will we count U-W-Z-V-X-Y and U-Y-X-V-Z-W as distinct? Should this be 10 and 60? or 120 and 120? Or something else?

And if we think about George’s process, is he arbitrarily making connections? If so, what happened when he reached U-V-W-X-U? Did he toss that out? Or are there decisions that we assume he is making along the way (He will not complete a 4 cycle, but a 3 cycle is ok).

As the question is written, we do not know.

Work, explained

Assume for the moment that the only possibility is one continuous loop through the six points. Think of the points sitting down around a circular table. We win if they sit U-V-W-X-Y-Z or U-Z-Y-X-W-V. But how many ways can they sit around a table?

Let’s solve an easier question.

  • Me and Kim sit around a table. She’s to my right, and to my left, no matter how we sit. That’s just one way to sit.
  • Now me, Kim and Laura sit around a table. I can be on Kim’s right, and Laura on Kim’s left. Or I can be on Kim’s left, and Laura on Kim’s right. That’s it. Two ways.
  • Now me, Kim, Laura, and McRib sit down. Seat Kim first. Laura can go across from her. Then it’s McRib left, me right, or McRib right, me left. That’s two ways so far. Laura gets up. Now McRib sits across from Kim. Me on Kim’s left, Laura on Kim’s right, or Laura on Kim’s left, me on Kim’s right. That’s two more ways. Finally, McRib stands, I sit across from Kim. McRib can sit on Kim’s left, with Laura on Kim’s right, or Laura on Kim’s left, McRib on Kim’s right. That’s two more ways, or six altogether

In general, we are plunking one person down, and then putting the others in order. I see the three examples above as 1×1, 1×2×1, and 1×3×2×1. If we want to generalize, the biggest number is one less than the number of people, so with n people our biggest number will be n-1 and we multiply 1×(n-1)×(n-2)×…×2×1 – or, in a formula, 1×(n-1)! or simply (n-1)!

These are called circle permutations.

How many pairs of triangles can we make out of 6 points?

  • ∆UVW ∆XYZ
  • ∆UVX ∆WYZ
  • ∆UVY ∆WXZ
  • ∆UVZ ∆WXY
  • ∆UWX ∆VYZ
  • ∆UWY ∆VXZ
  • ∆UWZ ∆VXY
  • ∆UXY ∆VWZ
  • ∆UXZ ∆VWY
  • ∆UYZ ∆VWX

It’s easier to use a formula than to make a list – but which formula? Problem is, every formula gives an answer – but none of them also tell you if you used the right formula. And that list? Not so long. And it’s pretty well organized. See if you can find a missing pair. I don’t think you can.

But now that we have the list, we can make a formula. Make one triangle, the second falls into place. Start with U, U has to be in a triangle. Then pick a second point (5 choices) and then a third (4 choices left) and that is almost the end, because the three remaining points will make the other triangle. But hold on: multiplying 5×4 we counted UWZ, but also UZW. And they are the same triangle. So divide our total, 5×4, by 2, and we get 10, which is the number of pairs in the list.

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