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Simple but not so simple

Classik

Classik

Member
At the end of a banquet 10 people shook hands with each other. How many handshakes will there be in total?
 
I'd say 5 if 10 people shook hands with each other then there is only 5 pairs of hands. :} unless of course you mean each person shook each other's hands then there would be 55 as stated earlier. :)
 
Meatballsub said:
I'd say 5 if 10 people shook hands with each other then there is only 5 pairs of hands. :} unless of course you mean each person shook each other's hands then there would be 55 as stated earlier.
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Goodness! You also go with 55?
 
If each person is to shake hands with all others then I'd have to go with the 90.

I don't remember my statistical math so I have to do it the long way.

Individuals A, B, C, D, E, F, G, H, I, J.
A+B, A+C, A+D, A+E, A+F, A+G, A+H, A+I, A+J = 9
B+C, B+D, B+E, B+F, B+G, B+H, B+I, B+J = 8
C+D, C+E, C+F, C+G, C+H, C+I, C+J = 7
D+E, D+F, D+G, D+H, D+I, D+J = 6
E+F, E+G, E+H, E+I, E+J = 5
F+G, F+H, F+I, F+J = 4
G+H, G+I, G+J = 3
H+I, H+J = 2
I+J = 1

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

Did I leave anyone out? I assumed nobody shakes hands with themselves.
 
Dang me, he's right. I shouldn't have had that '10' in my calculation.

My money's on WIP.
 
WIP said:
If each person is to shake hands with all others then I'd have to go with the 90.

I don't remember my statistical math so I have to do it the long way.

Individuals A, B, C, D, E, F, G, H, I, J.
A+B, A+C, A+D, A+E, A+F, A+G, A+H, A+I, A+J = 9
B+C, B+D, B+E, B+F, B+G, B+H, B+I, B+J = 8
C+D, C+E, C+F, C+G, C+H, C+I, C+J = 7
D+E, D+F, D+G, D+H, D+I, D+J = 6
E+F, E+G, E+H, E+I, E+J = 5
F+G, F+H, F+I, F+J = 4
G+H, G+I, G+J = 3
H+I, H+J = 2
I+J = 1

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

Did I leave anyone out? I assumed nobody shakes hands with themselves.
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This is actually correct if the person going down the line doing the hand shaking, leaves the "equation." If every single person has to take their turn going down the line and shakes the hands of the other 9, then the answer of 90 is correct.
 
WIP said:
If each person is to shake hands with all others then I'd have to go with the 90. I don't remember my statistical math so I have to do it the long way.Individuals A, B, C, D, E, F, G, H, I, J.A+B, A+C, A+D, A+E, A+F, A+G, A+H, A+I, A+J = 9B+C, B+D, B+E, B+F, B+G, B+H, B+I, B+J = 8C+D, C+E, C+F, C+G, C+H, C+I, C+J = 7D+E, D+F, D+G, D+H, D+I, D+J = 6E+F, E+G, E+H, E+I, E+J = 5F+G, F+H, F+I, F+J = 4G+H, G+I, G+J = 3H+I, H+J = 2I+J = 1Result 1+2+3+4+5+6+7+8+9 = 45Did I leave anyone out? I assumed nobody shakes hands with themselves.
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Great atttempt but...
 
Vanguard said:
This is actually correct if the person going down the line doing the hand shaking, leaves the "equation." If every single person has to take their turn going down the line and shakes the hands of the other 9, then the answer of 90 is correct.
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:sad:sad:shocked
 
counting the first shaker's shakes also that would then be 100 every time you shake some ones hand you shake your own.

So each person shakes his own hand 10 times.. ten times ten = 100 ...each person also shakes others hand 9 times 100 times 9 = 900 900+ 100 +1000 :)
 
reba said:
counting the first shaker's shakes also that would then be 100 every time you shake some ones hand you shake your own. So each person shakes his own hand 10 times.. ten times ten = 100 ...each person also shakes others hand 9 times 100 times 9 = 900 900+ 100 +1000
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Isn't she fantastic??? Wooooooow. Great attempt but...I'm also disappointed. Better luck;):toofunny
 
Classik said:
At the end of a banquet 10 people shook hands with each other. How many handshakes will there be in total?
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Of course I won't say anything about this being on Yahoo! back in 2007...though they used the term conference instead of a banquet! ;)
 
The answer is (10 choose 2) = 10 * 9 / 2 = 45. Each of the 10 people shakes hands with 9 others, but this counts every handshake twice, once for each person, so we need to divide 90 by 2.
 
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