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I am trying to calculate the number of treble bets for any given number of selections. A treble is a bet where 3 different selections in 3 different events are combined. So, selection a in event 1 selection b in event 2 and selection c in event 3 constitute a treble.

I can calculate the number of trebles for a given number of selections

4 selections = 4 trebles, 5 = 10, 6 = 20, but what I can't calculate, and need help with is the effect of a non runner. If there is 1 non runner say selection a, then each bet containing selection a becomes a double ie abc becomes a doulbe bc. If there are 2 non runners then each bet containing those non runners becomes a single ie abc becomes single c. If all three selections are non runners then the bet becomes void. I hope the following table illustrates more accurately what I mean. For 5 selections

sel nr  treble  double  single  void
5   0   10      0        0      0
4   1   4       6        0      0
3   2   1       6        3      0
2   3   0       3        6      1
1   4   0       0        6      4
0   5   0       0        0     0

Clearly there is a pattern but I cannot figure out how the pattern is arrived at, or how to calculate the effect of X non runners on treble bets of N selections. Please can anyone help. Many thanks.

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1 Answer

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Suppose that there are $n$ selections and $k$ non-runners.

  • A treble must contain only runners, so there are $\binom{n-k}3$ possible trebles.

  • A bet that contains one non-runner and two runners is demoted to a double; there are $k$ ways to choose the non-runner and $\binom{n-k}2$ ways to choose the two runners, so there are altogether $k\binom{n-k}2$ doubles.

  • A bet that contains two non-runners and one runner is demoted to a single; there are $\binom{k}2$ ways to choose the two non-runners and $n-k$ ways to choose the runner, for a total of $\binom{k}2(n-k)$ singles.

  • A bet is void if and only if it contains three non-runners; there are $\binom{k}3$ ways to form such a bet.

This gives you the complete breakdown. For computational purposes you may find it handy to rewrite these:

$$\begin{align*} \binom{n-k}3&=\frac{(n-k)(n-k-1)(n-k-2)}6\\ k\binom{n-k}2&=\frac{k(n-k)(n-k-1)}2\\ \binom{k}2(n-k)&=\frac{k(k-1)(n-k)}2\\ \binom{k}3&=\frac{k(k-1)(k-2)}6\;. \end{align*}$$

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Brian that is fantastic! Many thanks. I have struggled for days with that problem. And thank you for putting it so clearly. –  sila Oct 31 '12 at 18:01
    
@sila: You’re very welcome. –  Brian M. Scott Oct 31 '12 at 20:07
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