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
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answer #1
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answered by oldprof 7
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read it like a map for row x column
123456
24681012
369121518
4812162024
51015202530
61218243036
2006-09-29 16:38:24
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answer #2
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answered by D 1
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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
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answer #3
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answered by Helmut 7
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