| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1137.1 | | WONDER::COYLE | Only 48.8% of my former self! | Tue Oct 10 1989 16:40 | 23 |
| I do not really understand what you are trying to do; but I will
answer the question about finding the Number of Three Digit
Combinationsfrom a given number of Numbers.
Where the given number is N:
COMBINATIONS = N*(N-1)*(N-2)/6
Basicaly you take the nuber of possible first choices, times
the number of send choices, times the number of third choices.
This give the number of three digit Permutations. Since you
want combinations, persumably because the order of choices
wasn't important, you divide by the number of arrangements
of the three selected numbers. Which you get by having three
first choices times two second choices with the final choice
being already decided.
If you can better explain what you are really truing to do I will
try to help.
-Joe
|
| 1137.2 | Let me know if you need more info! | DNEAST::FREEMAN_KEVI | The Squeeky Wheel = Neglect | Wed Oct 11 1989 08:14 | 21 |
| Joe,
Sorry for not making myself clear. I indicated that each player was to
play against the other players ONLY once. I failed to indicate that
all players were to play against one another. For example using 9 players.
12312
45645 I get 123, 456, 789, 147, 258, 369, 159, 267, 348, 357, 168, 249
78978 by circling 3 digit combo's diag. up/down, and across.
By using your equation I get 84. Anyway, if I try my matrix with 11
players I get the same combo's as above and one other and the rest (16)
of two player matches. 1 vs 10, 2 vs 10 etc. That results in 12 three
player matches and 16 two player matches. If I randomly select combo's
I've achieved 15 three player matches for the same 11 players. Is
there a formula that will at least tell me the max. number of 3 digit
combo's for any value, and Ideally a formula that yields the combo's
ie first digit = X, second digit = X + 2, third digit = X + 4 or
something.
Regards, Kevin
|
| 1137.3 | 220 Possibilityies | WONDER::COYLE | Only 48.8% of my former self! | Wed Oct 11 1989 10:15 | 129 |
| When I go through the formulae I get 220 possible 3 player matches.
I have also listed them below. I suggest that you pick your matches
from the bottom group, then from the next group up (picking out
those that have not previously played) and so forth until you reach
the top.
-Joe
Here are all of the combinations with Player 12:
(55 Combinations)
12 11 10 12 10 9 12 9 8 12 8 7 12 7 6
12 11 9 12 10 8 12 9 7 12 8 6 12 7 5
12 11 8 12 10 7 12 9 6 12 8 5 12 7 4
12 11 7 12 10 6 12 9 5 12 8 4 12 7 3
12 11 6 12 10 5 12 9 4 12 8 3 12 7 2
12 11 5 12 10 4 12 9 3 12 8 2 12 7 1
12 11 4 12 10 3 12 9 2 12 8 1
12 11 3 12 10 2 12 9 1
12 11 2 12 10 1
12 11 1
12 6 5 12 5 4 12 4 3 12 3 2 12 2 1
12 6 4 12 5 3 12 4 2 12 3 2
12 6 3 12 5 2 12 4 1
12 6 2 12 5 1
12 6 1
Here are all of the combinations with Player 11 but without 12:
(45 Combinations)
11 10 9 11 9 8 11 8 7 11 7 6 11 6 5
11 10 8 11 9 7 11 8 6 11 7 5 11 6 4
11 10 7 11 9 6 11 8 5 11 7 4 11 6 3
11 10 6 11 9 5 11 8 4 11 7 3 11 6 2
11 10 5 11 9 4 11 8 3 11 7 2 11 6 1
11 10 4 11 9 3 11 8 2 11 7 1
11 10 3 11 9 2 11 8 1
11 10 2 11 9 1
11 10 1
11 5 4 11 4 3 11 3 2 11 2 1
11 5 3 11 4 2 11 3 2
11 5 2 11 4 1
Here are all of the combinations with Player 10 but without 11 and 12:
(36 Combinations)
10 9 8 10 8 7 10 7 6 10 6 5 10 5 4
10 9 7 10 8 6 10 7 5 10 6 4 10 5 3
10 9 6 10 8 5 10 7 4 10 6 3 10 5 2
10 9 5 10 8 4 10 7 3 10 6 2 10 5 1
10 9 4 10 8 3 10 7 2 10 6 1
10 9 3 10 8 2 10 7 1
10 9 2 10 8 1
10 9 1
10 4 3 10 3 2 10 2 1
10 4 2 10 3 1
10 4 1
Here are all of the combinations with Player 9 but without 10 through 12:
(28 Combinations)
9 8 7 9 7 6 9 6 5 9 5 4 9 4 3 9 3 2 9 2 1
9 8 6 9 7 5 9 6 4 9 5 3 9 4 2 9 3 1
9 8 5 9 7 4 9 6 3 9 5 2 9 4 1
9 8 4 9 7 3 9 6 2 9 5 1
9 8 3 9 7 2 9 6 1
9 8 2 9 7 1
9 8 2
Here are all of the combinations with Player 8 but without 9 through 12:
(21 Combinations)
8 7 6 8 6 5 8 5 4 8 4 3 8 3 2 9 2 1
8 7 5 8 6 4 8 5 3 8 4 2 8 3 1
8 7 4 8 6 3 8 5 2 8 4 1
8 7 3 8 6 2 8 5 1
8 7 2 8 6 1
8 7 1
Here are all of the combinations with Player 7 but without 8 through 12:
(15 Combinations)
7 6 5 7 5 4 7 4 3 7 3 2 7 2 1
7 6 4 7 5 3 7 4 2 7 3 2
7 6 3 7 5 2 7 4 1
7 6 2 7 5 1
7 6 1
Here are all of the combinations with Player 6 but without 7 through 12:
(10 Combinations)
6 5 4 6 4 3 6 3 2 6 2 1
6 5 3 6 4 2 6 3 1
6 5 2 6 4 1
6 5 1
Here are all of the combinations with Player 5 but without 6 through 12:
(6 Combinations)
5 4 3 5 3 2 5 2 1
5 4 2 5 3 1
5 4 1
Here are all of the combinations with Player 4 but without 5 through 12:
(3 Combinations)
4 3 2 4 2 1
4 3 1
Here are all of the combinations with Player 3 but without 4 through 12:
(1 Combination)
3 2 1
|
| 1137.4 | | ALIEN::POSTPISCHIL | Always mount a scratch monkey. | Wed Oct 11 1989 14:13 | 20 |
| Re .1, .3:
.0 isn't looking for all three-player combinations. It's just all
two-player combinations grouped into listings of three players -- each
item of three players represents three two-player combinations.
Re .2:
When it is possible to represent all the matches with groups of three,
there will be n(n-1)/6 such groups. But that is only possible when n
or n-1 is a multiple of three. When n has a remainder of two when
divided by three, it will not be possible to represent all the
two-player combinations with groups of three, because there are
n(n-1)/2 two-player combinations, and that won't be a multiple of three
which can be made into groups of three players, each group representing
three two-player combinations.
-- edp
|
| 1137.5 | may work for numbers greater that 9? | DNEAST::FREEMAN_KEVI | The Squeeky Wheel = Neglect | Thu Oct 12 1989 11:24 | 8 |
| Eric,
If I were to use the number 6 for instance (6*5)/6=5 and there are only
two varing combinations. ie 146,235 and the rest will have to be two
player combinations. It may prove true for values above 9 though I
haven't hacked it.
Kevin,
|