Dr Ian Plummer
Technical
How Many Clip Permutations
With four clips, twelve hoops and the peg how many different clip combinations are there? Discussions with various people have yielded wildly different numbers. The numbers appear to be in the tens of thousands. How could I find the answer and the math formula to prove the answer.
A nice question; answer 28,561.
This is my approach. Firstly, as you do not define it, I will assume that no balls have been pegged out. Any clip may be on one of thirteen items: hoops 1 to 6, 1b to rover and the peg. There are no interrelationships; any clip may be on any item irrespective of the other clips and, of course, the clips are distinguishable.
Consider two clips; the first clip may be on one of thirteen items, then the second clip may be on one of thirteen items independent of the first. Hence for any position of the first clip the second may be on one of 13 other locations: 13 positions of clip 2 if clip 1 is on hoop 1, 13 positions of clip 2 if clip 1 is on hoop 2, 13 positions of clip 2 if clip 1 is on hoop 3, ... This yields 13 x 13 combinations for two clips. For each of the positions of the first two clips the third clip can be in one of 13 positions ... By the same argument for four clips we get 13 x 13 x 13 x 13, i.e. 13^{4} = 28,561
I would qualify this by saying that I was never very hot at maths!
Adding Pegging Out
If the problem is extended to the number of permutations when balls can be pegged out, life gets a lot more difficult. The first thing to appreciate is that there will be different answers for handicap and level play.
Law 38. PEGGING OUT IN HANDICAP GAMES. The striker may not peg out the striker's ball in a stroke unless, before or during that stroke, the partner ball became a rover or an adversary’s ball has been pegged out. If he does so and removes the striker's ball from the court, Law 30 applies.
For level play when any rover ball can be pegged out at any time, we have to cope with the situation that in some cases two balls pegged out is the end of game (both from the same side) or they could be one ball from each side in which case the game continues.
Level play. There are now 14 locations for a clip: 12 hoops, the peg or the box. Considering just the red and yellow clips. We can have 14 positions for red and 13 for yellow without the game being over ...
Samir Patel:
Final conclusion  (15x13)^{2} = 38,025.
However, (15x13)^{2} simplifies to (14^{2}1)^{2} which gives the clue to the elegant solution:
Red has 14 possible clip positions. So does yellow, excepting the peg and peg option which can't happen while the game is still in progress.
Hence Red and Yellow have (14^{2}  1) clip positions.
Black and Blue are the same, so total combinations are (14^{2}  1)^{2}
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