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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.

N. F. Taussig
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Joe Black
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2 Answers2

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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 or the many equations for it ($!n$) on wikipedia https://en.wikipedia.org/wiki/Derangement.

Alexisz
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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

$$3 \times 3.$$

user2661923
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