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For example, I need number pairs, using numerals 1,2,3,4,5,6 and using each numeral twice. Each sequence should have six pairs. I know this likely doesn't make sense, but I'm math-illiterate, so please forgive me! One example would be 1&2, 3&4, 5&6, 1&3, 2&5, 4&6...thanks.

2006-09-29 09:15:36 · 3 answers · asked by mamaDee 3 in Science & Mathematics Mathematics

It's for a gift exchange...each # represents a person, and people can't give to themselves, or the person who is giving to them. Names will replace #'s later. Is there an easier way to do this?

2006-09-29 11:13:51 · update #1

3 answers

Not sure what you are asking. If you are asking for the number of possible combinations of six numbers (1 through 6) taken two at a time and allowing two of the same number, then:

Write out a six by six table with rows 1 through 6 and columns 1 through 6. The intersection of a row with a column represents the two digit combination. For example, where row 1 intersects column 4, you have the combination (1,4).

Since you allow duplicate numbers, like (3.3), the total number of diad combinations (two number values) is 6 X 6 = 36, which is the number of intersections (called elements) in the 6 X 6 table. If duplicate numbers were disallowed, the total number of combinations would be 36 minus the number of elements on the main diagonal (n = 6) where the duplicate numbers are found.

PS: Since you have six people and you can't (shouldn't) give a gift to yourself, take the main diagonal out of the table I described above. So that you'd have:

-1-2-3-4-5-6
1X o o o o o
2o X o o o o
3o o X o o o
4o o o X o o
5o o o o X o
6o o o o o X

So, from the first row, guest 1 can give to 2, 3, 4, 5, or 6; guest 2 can give to 1, 3, 4, 5, or 6, etc. Altogether, you have n^2 - n feasible combinations (sets of two) but (m, m) combiniations, giving to oneself, is disallowed.

2006-09-29 09:39:14 · answer #1 · answered by oldprof 7 · 1 0

read it like a map for row x column
123456
24681012
369121518
4812162024
51015202530
61218243036

2006-09-29 16:38:24 · answer #2 · answered by D 1 · 0 0

1,1..2,2..3,3..4,4..5,5..6,6
1,2..2,1..3,4..4,3..5,6..6,5
1,3..3,1..2,5..5,2..4,6..6,4
1,4..4,1..2,6..6,2..3,5..5,3
1,5..5,1..2,4..4,2..3,6..6,3
1,6..6,1..2,3..3,2..4,5..5,4

not as easy as it sounds.....

2006-09-29 17:06:41 · answer #3 · answered by Helmut 7 · 0 0

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