Hi Nauseated,
Your answer:
Consider this the same as unique combinations of 2 from integer set {n}
1 person, 0 pair
2 persons, 1 Pair
3 persons, 3 Pairs
4 persons, 6 Pairs
5 persons, 10 Pairs
6 persons, 15 Pairs
7 persons, 21 Pairs
8 persons, 28 Pairs
nC2 unique pairs
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You have found how many individual pairs can be made from 8 people.
That is NOT the question.
If there is three people, or any odd number of people in a room, they cannot all be paired of so the question would make no sense. There would be no answer.
You have said that if there are 4 people in a room there would be 6 pairs. This is not correct.
These are the the possible combinations
AB and CD
AC and BD
AD and BC
There are no other possible combinations. So there are 3 combinations.
If there are 6 people in a room then you are correct, there are 15 combination.
because
IF E and F are together there are 3 ways the others can be combined, as shown above.
IF E and A are together there are 3 ways the others can be combined, as shown above.
IF E and B are together there are 3 ways the others can be combined, as shown above.
IF E and C are together there are 3 ways the others can be combined, as shown above.
IF E and D are together there are 3 ways the others can be combined, as shown above.
So there is a total of 3*5 = 15 ways that 6 people can be paired off
Now there there is 8 people in a room A, B, C, D, E, F, G, H
G and H are together then there are 15 ways the others can be paired.
G and A are together then there are 15 ways the others can be paired.
G and B are together then there are 15 ways the others can be paired.
G and C are together then there are 15 ways the others can be paired.
G and D are together then there are 15 ways the others can be paired.
G and E are together then there are 15 ways the others can be paired.
G and F are together then there are 15 ways the others can be paired.
G and are together then there are 15 ways the others can be paired.
So there are 15*7 = 105 ways that the people can be paired
etc
If there are 2n people in the room then there will be 1*3*5*7*........*(2n-1) ways that they can be paired.
According to OEIS (Online Encyclopaedia of Integer Sequences)
https://oeis.org/search?q=1%2C3%2C15%2C105%2C&language=english&go=Search
The notation for this is (2n-1)!!
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How can you post a picture when the picture upload on the forum is not working properly ?
Thanks very much Alan for these instructions.
http://web2.0calc.com/questions/ven-diagram-need-help-for-questions#r3
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http://web2.0calc.com/questions/orchard-problem#r3
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https://web2.0calc.com/questions/simultaneous-equations-with-5-unknowns
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https://web2.0calc.com/questions/sin-5x-sin-3x
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http://web2.0calc.com/questions/complex-numbers_48321#r1
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These would be great practice for most students!
http://web2.0calc.com/questions/help_35979
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