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Basically, every girl is able to shake hands with 3 other people (6 people - 2 people beside them - themselves). So you have 6 people * 3 handshakes/person. Each handshake, however, is going to be counted twice, since there's one person on each end of the handshake. So the total amount of handshakes possible is:
Because it looked like a manageable number (6 girls) I thought brute force would be quick… 1 -> 3,4,5 2 -> 4,5,6 3 -> 1,5,6 (1 Double count) 4 -> 1,2,6 (2 Double count) 5 -> 1,2,3 (3 Double count) 6 -> 2,3,4 (3 Double count) My working on paper made more sense and didn't take long and I just striked-out double counts took me around 1:00
I’m not good at perm/combs but I redid it with that anyway: Question can be paraphrased to be how many pairs of girls except 6 combos: 6C2 = 6! / 2!4! = 15 15 combos in total 6 combos disallowed (cannot shake hands with girl on left covers the clause) Total combos: 15-6 = 9