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handshakes

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25 Jan 2006, 20:18
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If 10 persons meet at a reunion and each person shakes hands exactly once with each of the others, what is the total number of handshakes?
(A) 10â€¢9â€¢8â€¢7â€¢6â€¢5â€¢4â€¢3â€¢2â€¢1
(B) 10â€¢10
(C) 10â€¢9
(D) 45
(E) 36
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25 Jan 2006, 20:48
A for me..
The fisrt person can shake hands with 10.
Second can shake hands with 9
3rd with 8
....
so 10.9.8.7...1
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25 Jan 2006, 20:59
I think this problem is same as finding the number of combinations in which 2 people (out of 10) can be chosen at a time.

So it is 10C2 = 10!/(8!*2!) = 45.
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25 Jan 2006, 21:15
giddi's approach is good and simple))
D it is)) it can be drawn like this in order to make it clear:
1pers can shake hands with 2 3 4 5 6 7 8 9 10 pers
2-------------------------------3 4 5 6 7 8 9 10
3-------------------------------4 5 6 7 8 9 10
4-------------------------------5 6 7 8 9 10
5-------------------------------6 7 8 9 10
6-------------------------------7 8 9 10
7-------------------------------8 9 10
8-------------------------------9 10
9-------------------------------10
Total shakes))1+2+3+4+5+6+7+8+9=45
BTW Andy A is a trap answer))
It can't be 10 .One can't shakehands with himself))
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25 Jan 2006, 23:45
D. 45

9+8+7+6+5+4+3+2+1 = 45

Or

10C2 = 45
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26 Jan 2006, 11:37
We need to count the various tuples (0,1) , (0,2) and so on for the 10 people while eliminating the duplicates (0,1) and (1,0).

This can be done in 10C2 = 45 ways.

Another way to this is, if we count the first handshake between the first two people, we can count that as one pair and ignore the other.

P P Ignore Ignore Ignore Ignore Ignore Ignore Ignore Ignore

The total number of arrangements = 10!
Number of duplicates P =2 and Ignore = 8.

Thus, number of handshakes = 10!/2!*8! = 45.
26 Jan 2006, 11:37
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