Combinatorics with multiple conditions Sorry if the title is misleading/ flat out wrong.
Imagine 4 guys. They each have a dollar. Each person can give his dollar to someone else or keep it. Thus there are 4 choices each and 4x4x4x4 outcomes. Or n^n.
Say they can't save it. Now each has only 3 choices for 3x3x3x3 possibilities. Or (n-1)^n.
So they can save it, BUT at the end, nobody can have more than one dollar. This gives 4x3x2x1 or n!
How do you express both conditions? 
If you can't keep it and can't accept more than one dollar?
The first guy has 3 choices. Easy.
Now the second guy has either 2 or 3 choices depending on whether he received the 1st man's dollar. That's where I get stuck. What formula answers "how many endstates satisfy both conditions?"
 A: You are looking for the number of derangements of $4$ objects, that is, the number of permutations of these $4$ objects with no fixed point.
The problem is classical, and you will find very nice formulas for the number of derangements of $n$ objects in the article linked to.
For $4$ objects, you can find the number by careful enumeration. It will simplify things if you note that the number of derangements is $3$ times the number of derangements in which Person 1 gives her dollar to Person 2.
A: You have to count the dearangements of the set $[1,2,3,4]$. These are the permutations without a fixpoint. A nice formula for the number of dearrangements on $n$ numbers is $round(\frac{n!}{e})$ . Round means correct rounding. For $n=4$, we have $9$ possibilities.
The following PARI-program shows the possibilities for $n=4$ :
? q=0;for(j=1,4!,x=numtoperm(4,j);if(prod(k=1,4,k-x[k])<>0,q=q+1;print(x)));prin
t(q)
[2, 1, 4, 3]
[2, 3, 4, 1]
[2, 4, 1, 3]
[3, 1, 4, 2]
[3, 4, 1, 2]
[3, 4, 2, 1]
[4, 1, 2, 3]
[4, 3, 1, 2]
[4, 3, 2, 1]
9

