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Maths prob - Possible outcomes
How do I work it out?

 12:36 pm on Apr 15, 2014 (gmt 0)

Six races with 4 runner (runners named a.b.c.d) a,a,a,a,a,a or a,a,a,a,a,b or a,a,a,a,a,c or a,a,a,a,a,d Im not sure if thats clear.

Whats the answer ?



 12:56 pm on Apr 15, 2014 (gmt 0)

I think it is 4 (runners) to the power of 6 (number of races) - so 4096.


 1:42 pm on Apr 15, 2014 (gmt 0)

That makes sense thanks


 2:55 pm on Apr 15, 2014 (gmt 0)

Hi there Essex_boy,
my arithmetic gives 144 as the answer. ;)

Four runners with short, easy to remember, monosyllabic
names have 24 possible finishing orders if they decide to
compete against each other in a race.

If they are unsatisfied with the result of the first race
and decide to do it again, then another 24 possible
finishing orders can be added to this total.

These four guys, though, are devoid of sensibility and
continue this idiotic pursuit a further four times, which
adds up to 144 possible finishing orders by my reckoning.

My arithmetic, of course, may be totally wrong, as I am...

birdbrain :(


 4:32 pm on Apr 15, 2014 (gmt 0)

Is Usain Bolt running in the races? If so, that limits the realistic outcomes quite a bit and more likely just 12 outcomes for each race. :)


 5:07 pm on Apr 15, 2014 (gmt 0)

Hi there LifeinAsia,

If Usain Bolt was foolish enough to run in a race with members
of the alphabet and assuming that he would arrive in first place
over all distances, then there would be just 6 outcomes for
each race....
  1. Usain Bolt, b, c, d
  2. Usain Bolt, b, d, c
  3. Usain Bolt, c, b, d
  4. Usain Bolt, c, d, b
  5. Usain Bolt, d, b, c
  6. Usain Bolt, d, c, b

I feel sure that the alphabeters would realise that resistance was
futile and end proceedings at that juncture. ;)



 5:10 pm on Apr 15, 2014 (gmt 0)

Correct- 6. It should be illegal for anyone to attempt math problems before being sufficiently caffeinated for the day...


 6:06 pm on Apr 15, 2014 (gmt 0)

Birdbrain - Thats incorrect.


 6:23 pm on Apr 15, 2014 (gmt 0)

Whats the answer ?

What's the question? Are we talking about relay races, where the OP simply showed one possible starter for each of the six (leaving 3! = 6 possible configurations for the remaining three legs of each race using the information given)?

Anyway, where are you finding these four people who all perform identically in all legs of a relay, regardless of how many races they're already run?


 6:33 pm on Apr 15, 2014 (gmt 0)

Hi there Essex_boy,
what is incorrect, my interpretation of your question or my solution?

If you meant the total combination of six races then that would be...


...or 24 to the power of 6. ;)



 7:53 am on Apr 16, 2014 (gmt 0)

What about photo-finish where two or more have equal times?
Or, a runner or more retire before the race finishes?
Or, one or more fails a blood test?


 4:17 pm on Apr 16, 2014 (gmt 0)

birdbrain, you were right the 1st time, 144.

You can't do 24 to the power of six, this would include race results like d,b,b,b and b,b,b,b.

Lame_Wolf, you're a trouble maker, and I love it!


 4:38 pm on Apr 16, 2014 (gmt 0)

I'm with lucy24
The races could be marathons, or 100m sprints, or a combination of any. There could be injury, or sabotage, false starts and disqualification, or one of the runners doesn't show up because he's eaten too many pies, and, the runners may be old and young, fit and unfit, suited to a sprint or to a marathon, etc.

In other words, the outcome is not a single figure because of the number of possibilities.


 5:26 pm on Apr 16, 2014 (gmt 0)

I was a History major and don't have a clue.

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