Thread: counting handshakes

Near the end of the party, everyone shakes hands with everybody else. A straggler arrives and shakes hands with only those people whom the straggler knows. Altogether, 68 handshakes took place. How many people at the party did the straggler know?

Originally Posted by ihavvaquestion

Near the end of the party, everyone shakes hands with everybody else. A straggler arrives and shakes hands with only those people whom the straggler knows. Altogether, 68 handshakes took place. How many people at the party did the straggler know?

Assume that "everyone shakes hands with everybody else" means "everyone shakes hands with everybody else exactly once." Same for the straggler's hand shakes.

Let n represent the number of people at the party (not including the straggler), and k represent the number of people the straggler knows.

The first part involves C(n, 2) handshakes, that is, for every pair of people there is one handshake, and there are C(n, 2) ways to pair the people.

The second part is simply k.

So C(n, 2) + k = 68.

Now we might be suspicious that the problem does not give enough information because there are two unknowns and one equation. BUT we know 0 <= k <= n. So take a look at:

C(11, 2) = 55
C(12, 2) = 66
C(13, 2) = 78

The maximum value of C(n, 2) + k for n = 11 is 66, too low.

The minimum value of C(n, 2) + k for n = 13 is 78, too high.

Therefore n = 12, and k = 2.