SOLUTION: 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 handsh

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Question 38562: 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?
Found 2 solutions by AnlytcPhil, fractalier:
Answer by AnlytcPhil(1807)   (Show Source): You can put this solution on YOUR website!
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?

===============================================================================

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
Answer by fractalier(6550)   (Show Source): You can put this solution on YOUR website!
This can be done a couple of ways...
Let me explain it this way.
The first of 8 people can shake hands with 7 people.
The second of 8 people can shake hands with 6 people (he has already shaken hands with the first).
And it goes on like this, so that the number of total handshakes for the 8 people is
7 + 6 + 5 + 4 + 3 + 2 + 1 = 28 handshakes
leaving 4 handshakes by the ninth person, which makes the total of 32.
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