Finding the amount of possibilities to split two parties in pairs I have two parties. Each party contains $3$ people. I'm trying to figure out how many possibilities there are to divide those two parties into pairs so each pair contains a person from party A and a person from party B. 
I think the answer is $6$. But I'm trying to prove it. First of all, I put the three people from party A in a row in $3!$ options. I also put the three people from party B in a separate row in $3!$ options. Each pair will be in the people by indexes (first pair contains the first person from the first row and the first person from the second row and so on).
But now we need to divide by something. How should I continue this idea?

You are correct that there are six ways to match people from party $A$ with people from party $B$ in pairs.
Line up the people in party $A$ in some order, say alphabetically.  There are three ways to match a person from party $B$ with the first person in line, two ways to match one of the remaining people from party $B$ with the second person in line, and one way to match the remaining person from party $B$ with the third person in line.  Hence, there are $3! = 6$ ways to pair people from party $A$ with people from party $B$.
To continue with your idea, match the person from party $A$ with the corresponding person in party $B$.  Notice that doing so counts each possible pairing $3!$ times, once for each of the $3!$ ways we could arrange the same set of pairs in the two lines.  Hence, there are
$$\frac{3!3!}{3!} = 3!$$
ways to pair people from party $A$ with people from party $B$.
