You should be able to solve this

You should be able to solve this.

>some of the people shook hands with some of the others
There is no solution.

Fuck Mrs. Smith

The key is that every single person gave a different answer. But i'm too lazy to think rn...

Still, Mr. Fuckface could be any of those answers. There isn't a unique solution.

Obviously the answer is 4

>Obviously
Explain it then you fucking retard.

You can't figure it out by yourself brainlet? OK I'll spoonfeed you the answer. Start by noticing that one can only have between 0 and 8 handshakes since there are 10 people and you can't shake your spouse's or your own hand. But that means at most 9 people can have a unique number of handshakes, which is perfectly fine since the question states that only the people Mr. Smith questioned have unique numbers. Mr. Smith didn't say anything about his own number. So let's say one of the other couples contains the person with number 8. Then his spouse must be number 0 because everyone else shook hands with number 8. Similarly, the person whose number is 7 has spouse with number 1 since everyone else shook hands with number 7 and number 8. This continues until we get to the couple with 5 and 3. Then the last couple has 4 and 4, from shading hands with 8, 7, 6, and 5. This must be Mr and Mrs Smith since 4 is repeated. Therefore Mrs Smith has 4 handshakes. QEDp

Where in the problem was it said you can't shake a person's hand if you shook their spouse's?

Now an idiot like you probably gave up immediately since you assumed there was no way to distinguish between the couples, which was wrong. A smarter brainlet would have looked at the number of unique handshakes and given up because they assumed Mr. Smith counted. You were given the answer and you didn't even try to figure it out. You're lower than a brainer.

Where in my explanation did I claim that? Geez you're dumb.

A solution exists: W1 shakes every possible hand (8), H1 shakes no hands (0).

W2 shakes every hand but H1's (7). H2 shakes no hands but W1's (1).

W3 shakes every hand but H1 and H2 (6). H2 shakes no hands but W1 and W2's (2)

W4 shakes every hand but the previously mentioned husbands (5). H3 shakes no hands but the previously mentioned wives (3).

W5 and H5 both shake with W1-W4.

Shakes are therefore {0,1,2,3,4,4,5,6,7,8} and Mr. Smith must shake 4 in order for everyone else to be able to provide a different answer.

The solution is unique: draw the fuckin graph.

A little late to the party.

>sure, you didn't claim it
>but you sure as hell used your logic as if you did
What are you?

No retard. You didn't understand the proof at all. The reason the spouse of number 8 must be number 0 is because everyone else shook hands with number 8, which means their number is 1 or greater, so they can't be 0. Only 8's spouse can be 0. It has nothing to do with some assumption.

Well Veeky Forums?

8 times
Mrs Smith shakes hands with everyone besides her husband
4 other couples = 8 people

>how many times .... shake someone's hand
Depends on on how long they shook hands (up-down)

Viz. shake someone's hand =/= shake hands with everyone.

Its ok, I'm a retarded brainlet whose has no idea what I'm talking about, please ignore me

Thats correct.