Here's the question you clicked on:
sritama
Three married couples arrange a party. They arrive at the party one at a time, the couples not necessarily arriving together. They all, upon arriving, shake the hand of everyone already there, except their own spouse. When everyone has arrived, someone asks all the others how many hands they shook upon arriving, and gets five different answers. How many hands did he himself shake upon arriving?
?? five different answers ??
The answer is he shook 2 people's hands.
@FoolFor, how could u get the answer?
must be six .. because there are 3 couples .. and this seventh person is an outsider
This is a very old combinatoric puzzle.
but i there were 3 couples and they shook each other hand .. except of their own spouse ... they they must have shaken hand 4 times ... each of them ...but why 5 different answers would come??
@JoshDavoll if the person was insider .. then answer must be 4
he asked 5 people means including him, there were 6. that means he was the last to arrive. so, he shook 4. he didnt shake his own hands definitely, and definitely not with his spouse.
yeah, if he was the insider as you say.
i have a solution, gimme a sec
@FoolForMath is probably correct.
he asked them after all this happened, i think you misunderstand
number the couples 1 , 2 , 3
so m1 is male from couple 1 , and f1 is female from couple 1
if he asked without shaking hand .. and the pair couple shook hand 3 times while the unpaired couple shook 4 times
first to arrive is m1, he shakes 0 peoples hands second is m2 he shakes 1 third is m3 he shakes 2 fourth is f1 she also shakes 2 fifth is f2 she shakes 3 sixth is f3 she shakes 4 the asker is either m3 or f1 , and they get 5 different answers
Ah i get it now, thanks for explaining it :)
how many hands ON ARRIVING?? i think i understood the question now.
of course we haven't checked thats the only solution, but you can change around the numbers to get a solution that looks different but is the same