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Mon 13 Sep, 2004 07:30 pm
I need help with this riddle plz.
Reccently, I attended a small get-together. I counted the number of handshakes that were exchanged. there were a total of 28. Everyone present shook everyone else's hand only one time.
How many people were present and how did you get the answer.
I can't figure it out.
It takes two people to make up a handshake. You want to know N such that N choose 2 (combinations of N things taken 2 at a time) equals 28. The formula for N choose 2 is:
N*(N-1)/2
7
you have to factor the possible combinations minus one combination per guest i.e.
guest 1 shook with guests 2,3,4,5,6, 7 for 6 possible combinations
guest 2 shook with 3,4,5,6,7 you dont count 1 cuz they only shake oncewhich gives you 5 combinations
each guest reduces the combinations by one add all the combinations for each guest to get 28
Guest combination
1 6
2 5
3 4
4 3
5 2
6 1
7 0
soe each guest shakes
You're method is correct, except that adds up to 21.
help!!
8 people
( A to H )
A shakes hand with 7 people ( B - H) and
B with 6 people ( C - H) and
C with 5 people ( D - H ) and
D with 4 people ( E - H ) and
E with 3 people ( F- H ) and
F with 2 people ( G and H ) and
G with the last 1 person that is
H
Total number of handshakes: 28
7+6+5+4+3+2+1 = 28
And everyone would have shaken the hand of everyone else,once.
Do I make sense? Lololol.
thanks marker for correcting my math error! I guess it was late, i never double checked the math
there was 8 people at the party