Number of Permutations

[adam2308]

What formula can I use to work how many permutations there that a snooker match can be a given score? Bearing in mind the losing player can not win the last match.

e.g. For Player A to beat Player B in a first to five match, 5 frames to 1. I know that there are five permutations of this as Player B needs to win one of the six frames played but he can not win the sixth as Player A would have already won the five required to win the match.

This is easy enough to work out but when it gets to Player A to win 5-4 it becomes more complexed and very tedious writing down all the permutations! I can't even face the thought of working out the number of permutations for a first to 18 match to finish 18-17 to a given player!!!

Thanks for any help,
Adam.

[shg]

Well ...

If Player A wins 5-1, then the last point is (per force) a win. Among the other 5, B wins one, so there are =COMBIN(5, 1) = 5 ways that could happen.

To win 5-4, there are 8 rounds other than the last, 4 of which are won by player B, so it could happen in=COMBIN(8, 4) = 70 ways.

To win 18-17, there are 34 rounds other than the last, 17 of which are won by player B, so it could happen in=COMBIN(34, 17) = 2,333,606,220 ways.

[daddylonglegs]

Do you just need the number, or a list of the combinations? If it's the former you can use COMBIN function in Excel, e.g. winning number in A2, losing number in B2, e.g. 5 and 1, then in C2 this formula will give your result

=COMBIN(A2+B2-1,A2-1)

Edit: shg's suggestion effectively uses

=COMBIN(A2+B2-1,B2)

which is better because it's shorter.....but it'll give you the same result

[shg]

Or =COMBIN(A2+B2-1, B2)

[daddylonglegs]

I just edited to that effect.......

[shg]

Just to explain (to Adam, not DLL!)...

Any combination of n Choose m is the same as n Choose n-m, so, for example, 10 Choose 3 is the same as 10 Choose 7.

Which makes sense, because you can consider 10 Choose 3 equally as choosing the 3 to include or the 7 to exclude.