From Enriqueta's post, it does not seem like the order of selection matters. Picking John and Kelly is the same as picking Kelly and John. For 2 pairs to be different, they must not include the same 2 children. This matters
Amising's solution would be correct only if order of selection mattered.
For instance, suppose there are 3 children and we want to figure out how many pairs are possible. Amising's solution would be:
>> choose the first child from 3 options.
>> choose the second child from 2 options.
>> a total of 6 pairs.
Actually there are fewer than 6 pairs. With 3 children A, B, and C the only possibilities are (A,B), (A,C) and (B,C). This is because (A,B) and (B,A) is the same pair. The solution with 3 children to pick from is 3, not 6.
This is a straightforward combination problem. How many ways can we group 100 kids into pairs is 100C2 (choose 2 from 100, order is irrelevan). This is 100!/(2!98!) = 4,950 pairs possible
-Patrick
