Question 38562
<pre><font size = 3><b>The 8 people at a party shook hands exactly once with each of the others before
the 9th arrived. The 9th person then shook hands with some of these 8 people. A 
total of 32 handshakes took place. With  how many people did the 9th person
shake hands?

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Before the 9th person arrived there were "combinations of 8 things taken 2 at a
time".  It is combinations, not permutations, because the order in which they
stand to shake hands does not matter.

C(8,2) = 8!/(2!6!) = 28 handshakes before the 9th person arrived

After the 9th person finished shaking hands, there were 32 handshakes
altogether.

Therefore the 9th person shook hands with 32 - 28 or 4 people.

Edwin
AnlytcPhil@aol.com</pre>