Question: I'm wondering if there is a way to calculate arrangements for this problem without enumerating all possibilities.

There are 4 work-pieces that 4 guys A, B, C, D work on during a single day. At the end of the day each person transfers his assigned work-piece to another person. Each person works on only 1 work-piece during a day. How many arrangements of transfer are possible?

E.g. A can transfer his workpiece to C (AC), or A can transfer it to B, or A can transfer it to D, but A can't transfer it to A.

One arrangement is:

```
AC (A transfers to C) 
BD (B transfers to D) 
CA  
DB

```

For the purposes of this question the order of transfer (e.g. whether AC is written in line above BD) doesn't matter, and the above arrangement is considered same as: `CA, BD, DB, AC`.

Answer: 
These are all the permutations of a set with 4 elements without fixed points. For a set with $n$ elements see: [Fixed-point-free permutations](https://math.stackexchange.com/questions/41949/fixed-point-free-permutations) or the many equations for it ($!n$) on wikipedia <https://en.wikipedia.org/wiki/Derangement>.
Alternative approach:

Person-A can transfer to anyone of three people.

So, there will be an initial factor of $(3)$.   

Now, you can assume, without loss of generality, that Person-A transfers to Person-B.

Now, there are 3 people that Person-B might transfer to:

* If Person-B transfers to Person-A, then the transfers are set, since Person-C and Person-D must then exchange.
* If Person-B transfers to Person-C, then Person-C must transfer to Person-D (or else no one is transferring to Person-D).
* Similarly, if Person-B transfers to Person-D, then Person-D must transfer to Person-C.

Therefore, if you know that Person-A transferred to Person-B, then there are $(3)$ possible ways that the transfers can be completed.

Therefore, the total number of satisfying groups of transfers is

$$3 \times 3.$$