MHB How Many Hands Did You Shake at the Party?

[Ackbach]

Here is this week's Problem of the Week; you might notice a similarity with the University POTW # 147, two weeks ago, and there are. But there are also some subtle differences which completely change the solution.

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Suppose you and your husband attended a party with three other married couples. Several handshakes took place. No one shook hands with himself (or herself) or with his (or her) spouse, and no one shook hands with the same person more than once. After all the handshaking was completed, suppose you asked each person, including your husband, how many hands he or she had shaken. Each person gave a different answer. How many hands did you shake? How many hands did your husband shake?

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[Ackbach]

This problem is from Introductory Graph Theory, by Gary Chartrand.

Congratulations to Fallen Angel for his correct solution, which you can see below:

Spoiler

There are eight people at the party, namely $\{\text{Me},\text{My spouse},m_1, m_2, m_3 , f_1 ,f_2, f_3\}$ , where $m_i -f_i$ is a couple; each one has $6$ possibles hand-shakes and you have $7$ different answers, so the answers should be $0,1,\dots ,6$.

Assume $m_{1}$ has shaken $6$ .

Now $f_{1}$ is the only one that can shake no one (How ill-mannered! ).

If $m_{2}$ has shaken $5$, then $f_{2}$ is the only one that can shake once.

And if $m_{3}$ has shaken $4$, then $f_{3}$ has shaken twice.

So my spouse and I have shaken $3$ each.

Notice that this is totally symmetric, and every $m$ can be replaced by $f$ and vice-versa, and every $i$ by every $j$; but my spouse and I can't change with another couple because then there will be two identical answers between the guests.