# Puzzle Probability - How many options

#### Dramine

##### New member
Hello fellow Mathematics Members. After finishing school I have spent some time working on mathematic exercises to enhance my knowledge. I have stumbled upon this exercise I cannot solve. If anyone could give me a step by step explanation (leading up to the solution), it would be great. I have spent hours on it already (including scrolling through websites) but I haven't found a clue on how to solve it:

All 4 people have the same empty (not filled) Puzzle with (consisting of) 4 different parts and they now each write their own name on all of their 4 parts. They now give one part to the other person so that in the end everyone has one set with all 4 (different) names on them. Then how many ways are there to do this?

I would really appreciate a fast answer

I wish you all a nice day/night!

Sincerely Dramine

#### lev888

##### Full Member
By puzzle you mean jigsaw puzzle? All pieces are different?

#### Dramine

##### New member
By puzzle you mean jigsaw puzzle? All pieces are different?
Yes exactly. And all four jigsaw parts are different. As far as I know the jigsaw puzzle copy everyone has, should be same though (Because it has to form one jigsaw puzzle when everyone exchanged parts).

#### Jomo

##### Elite Member
I would really appreciate a fast answer
Are you really serious? You want us to the problem for you? On top of that you want it fast. The title of the website is not free math solutions but rather free math help. It is like a game, you submit a problem that you can't do, show the work that you tried to solve the problem and then some tutor will guide you to the correct solution. It really is fun and I truly hope that you give it a try.

I'll even give you a hint. Suppose the four people are A, B, C and D, and the 4 pieces are called w, x, y and z. What might a listing of what the 4 people have look like. If you know what the form of what something should like you might be able to count all the combination.

Consider A has w from A (Aw), Bx, Cx, Dz and B has .... and C has and D has ..... This will yield just one result.