Advanced Correct Score Permutations - Help

[sendandreturn]

Hi, I understand and know the amount of doubles in a standard ten fold etc, but what I would like help with is how to calculate the amount of trebles in the below for example where there's X number of scores per match, eg... Match 1 : 1-0, 2-0, 2-1, 3-0 Match 2 : 1-0, 2-0, 2-1, 3-0 Match 3 : 1-0, 2-0, 2-1, 3-0 Match 4 : 1-0, 2-0, 2-1, 3-0 If I had 5 scores each for 2 matches, that's 25 doubles...but that's as far as I can work it out thanks... There used to be a great illustration on a site called www.1x2expert.com but thats been shut down.

[wiw4]

Re: Advanced Correct Score Permutations - Help I'm thinking 16*12*8/1*2*3 = 256 trebles

[clay747]

Re: Advanced Correct Score Permutations - Help To work out the number of combinations is:-

nCr = n! / (n-r)!r!

where n is the number of objects and r is the number of picks. (! means factorial.) So you have 16 scores and want to pick trebles, 16!/(16-3)!3! = 560. But with correct scores, many of your trebles will be illegal as they involve two or three scores from the same match. So you need to subtract the number of illegal trebles from 560 to find the correct answer. Each line has 4 illegal trebles(4!/(4-3)!3! = 4) and 6 illegal doubles(4!/(4-2)!2! = 6). So there are 16 illegal trebles and 24 illegal doubles in total. The illegal doubles will makeup 288 illegal trebles (24*12) plus the 16 illegal trebles gives, 288+16 = 304 illegal trebles. So, 560 - 304 = 256 possible trebles.

[slapdash]

Re: Advanced Correct Score Permutations - Help If you have the same number of scores per match (say X scores per match, and m matches), then I think the easiest way to do it is: First, calculate the number of ways to choose the three matches involved in a treble: this is just mC3 = m!/(m-3)!3!, as in clay747's post. Then, for each possibility for the 3 matches, there are X^3 possibilities for the three scores (X possibilities for the first, times X possibilities for the second, times X possibilities for the third). So in total there are X^m times mC3 trebles. If m=4 and X=4, as in your example, this is 4^3 times 4C3 = 64x4 = 256. Which agrees with everybody else! :nana If you have different numbers of scores for different matches, then it's messier.