(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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