another logic puzzle

[desh]

This came up today. I'm not 100% sure the answer I have in mind is unique, but I think it is.

You and your spouse invite 4 other couples over for a dinner party. During the course of the party, many people shake hands with other people, except that no one shakes hands with their partner or themselves. At the end of the party, you realize that each of the other 9 people (that is, everyone but you) shook a different number of hands from each other. How many different hands did you shake?

Again, the answer doesn't involve tricks, like some people shaking with both their left and right hands or shaking the hand of someone not at the party. All comments screened for now, like before. Questions and bad guesses will be unscreened sooner; good guesses and right answers will be unscreened later. No googling allowed! Comments now contain spoilers.

Comments

[desh]

Nope, that can't work. (Sure, socially you'd probably shake 8 hands, but then you can't end up with the condition that all 9 other people have different handshake numbers.)

[intangiblehugs.livejournal.com]

so each of the 9 other people shook a different number of hands from you and from each other? or just from each other?

rephrased: did you shake the same number of hands as anyone else?

[desh]

Just from each other.

[erin.livejournal.com]

I'm trying to read my book on clouds and for some reason this puzzle keeps coming up in my head.

You could shake hands with up to 8 people.

So that's a 0-8 range - a total of 9 different "numbers"

There are 10 people at the party. 10 is greater than 9. So it's not even possible for everyone to shake a different number of hands, is it?

The only other possibility is that for some reason you shook Bob's hand twice. I don't know, maybe you liked Bob.


Like this:

Couple A1 - 8
Couple A2 - 7
Couple B1 - 6
Couple B2 - 5
Couple C1 - 4
Couple C2 - 3
Couple D1 - 2
Couple D2 - 1

(that's 4 of the 5 couples)
Couple E1 - 0

And Couple E2 is an impossibility because there aren't any "different" numbers left.

So say Couple E2 is me, the host. I can shake anywhere from 0-8 hands and the other 9 will still be different from each other (not from me, but from each other).

I googled this - just now - and it appears the answer is 4. Which doesn't make sense because in all the solutions the assumption is that each couple shook a total of 8. The original problem doesn't state this nor does it state that the host's spouse absolutely shook someone's hand.

So what am I missing? :P

[desh]

See ag5j's answer below (his middle paragraph).

The reason your answer doesn't work: Who are the 8 people that A1 shook hands with? Not A2 and not A1 and not E1...so there aren't even 8 people left. The zero has to be A1's partner.

The "each couple shakes a total of 8" thing ends up being a consequence of the rules of the problem, not something you have to assume going in. And if you don't see this right away (as I didn't and you didn't, but some other commenters did), then it's a really hard problem.

[erin.livejournal.com]

ohhh. That make sense now. Thank you :)