Nope! This does not consider that no 2 of the 25 numbers are in a SET more
than once.
This brings the number of cases down substantially.
I was surprised that I was able to find only 4 solutions:
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 25
1 6 11 16 21
2 7 12 17 22
3 8 13 18 23
4 9 14 19 24
5 10 15 20 25
1 7 13 19 25
2 8 14 20 21
3 9 15 16 22
4 10 11 17 23
5 6 12 18 24
1 8 15 17 24
2 9 11 18 25
3 10 12 19 21
4 6 13 20 22
5 7 14 16 23
best, paul
At 08:50 AM 3/9/01 -0600, you wrote:
>(25 choose 5) x (20 choose 5) x (15 choose 5) x (10 choose 5) / 5! ways of
>ordering the piles.
>
>= 25!/(5!^6) = 5,194,672,859,376
>
>as a formula n!/(p!(s!^s))
>n = number of cards
>p = number of piles
>s = number of cards in piles
>
>From: [EMAIL PROTECTED] (paul smith)
>Reply-To: [EMAIL PROTECTED]
>To: CF-Talk <[EMAIL PROTECTED]>
>Subject: OT: Combinations (Weekend Recreation)
>Date: Sat, 03 Mar 2001 14:55:27 -0800
>
>I seem to recall a mathematics buff lurks here.
>
>I have 25 numbers: 1 thru 25
>
>I want to create 5 GROUPS where each contains 5 SETS of 5 numbers,
>and where EACH group contains ALL 25 numbers.
>
>For example, one such group is;
>
>1 2 3 4 5
>6 7 8 9 10
>11 12 13 14 15
>16 17 18 19 20
>21 22 23 24 25
>
>where SET 1 is: 1 2 3 4 5,
>
>SET 2 is: 6 7 8 9 10, etc
>
>How many GROUPS are there where no 2 numbers are in any SET more than once?
>
>What are these GROUPs?
>
>best, paul
>
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