SOLUTION: If 10 students go hiking in pairs, how many different pairs of students are possible?

Question 1110182: If 10 students go hiking in pairs, how many different pairs of students are possible?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
this is the same as asking how many sets of 2 can you get out of a set of 10, where order within each set of 2 is not important.

that would the combination formula of c(10,2) = 10! / (2! * 8!)

that comes out to be c(10,2) = 45.

let each student be numbered from 0 to 9 and the possible pairings are shown below.

0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9
1,2 1,3 1,4 1,5 1,6 1,7 1,8 1,9
2,3 2,4 2,5 2,6 2,7 2,8 2,9
3,4 3,5 3,6 3,7 3,8 3,9
4,5 4,6 4,7 4,8 4,9
5,6 5,7 5,8 5,9
6,7 6,8 6,9
7,8 7,9
8,9

order not being important within each set of 2 means that the set of (1,0) and the set (0,1) are considered to be the same set and are therefore not counted as 2 separate sets.

that is why you only set the set (0,1) and do not see the set (1,0).

if order within each set of 2 was important, then you would have to see the set (0,1) and the set (1,0) as 2 separate sets.

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