Herm and his 12 followers have created a not-so-secret ``handshake'' that involves three people. The first person (the hook) initiates the greeting by casting an invisible fishing line, and the second person (the bait) walks toward the first person as if the ``bait'' is being reeled via the invisible fishing line. When the bait is about to meet the hook, they both go up with hand as if to perform a ``high five.'' At that point, however, a third person (the bandit) steps in to ``steal'' the ``high five'' from the hook.
Within Herm's group, how many different handshakes are possible?
+4 You are looking for the number of permutations when you pick three things from a group of thirteen.
So
\({n! \over (n-m)!}\) where n=number of things in group and m=number of things you want to pick
\(x = {13! \over (13-3)!}\)
\(x = {13! \over 10!}\)
13*12*11
1716
Why does it work like that?
--->
Which one is going to be the bait?
One of these 13 people.
Which one is going to be the hook?
One of these 12 people that are not the bait already
Which one is going to be the bandit?
One of the 11 people left.
So you have 13*12*11=1716 possible handshakes
