Permutations and Combinations question There are 4 mice (A, B, C and D) and 5 exits each mouse has an equal chance of exiting any of the exits and which exit they choose does not depend on the other mice
 Total no. of ways mice can exit = 5^4=625
But when I went through each of the different cases I got a different total no. of ways. Can someone please tell me what I am doing wrong?
Case 1: All mice exiting 1 exit
5C1 = 5 possible ways
Case 2: 3 mice exiting from one and the other from a different
5 possible spots for the lone mouse, 4 possible spots for the other mice, 4 ways of picking the lone mouse
5.4.4=80
Case 3: 2 mice exit 1 exit and the other 2 exit together from a different exit
5 possible spots for the first pair, 4 possible spots for the second pair and can pick the first pair in 4C2 ways
5.4.4C2=120
Case 4: 2 mouse exit from one exit and each of the others exit alone
Pair can be chosen in 4C2 ways, 5 possible spots for pair to exit, other 2 can exit in 4P2 ways
4C2.5.4P2=360
Case 5: all 5 exit separately
First can exit in 5 ways, 2nd in 4 ways, etc.
5.4.3.2=120
But this total is 5+80+120+360+120=685 and not the 625 which is what the total number of ways should be
 A: the error is in Case #3.  
You neglect a symmetry in choosing the exits.  Choosing Exit A followed by Exit C is the same as choosing Exit C followed by Exit A, so you have to divide by $2$.  That changes your calculation from $120$ to $60$, which reconciles the counts.
A: If you insist, I'll find out the errors in your count, but there is a very simple way to solve it:
Each mouse has the choice of going out through any of the 5 exits,
thus # of ways = $5\times 5\times 5 \times 5  = 5^4 = 625$
PS
Since you insist, the corrections are:
Case 3: 2 mice exit 1 exit and the other 2 exit together from a different exit
"5 possible spots for the first pair, 4 possible spots for the second pair and can pick the first pair in 4C2 ways"
Spots for the 2 pairs are ${5\choose 2} = 10, \text{not} 5\times 4$,
You are double counting by taking A going out through gate 1 and B going out through gate 2 as different from B going out through gate 1 and A going out through gate 2. The correct count for this cases is thus 60, not 120.
