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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

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    $\begingroup$ The error is in Case #3. Choosing Exit A, then Exit C is the same as choosing Exit C then Exit A, so you must divide by $2$. That changes the $120$ to $60$. $\endgroup$
    – lulu
    Sep 6 '15 at 12:02
  • $\begingroup$ @lulu: this should be put into an answer. $\endgroup$
    – robjohn
    Sep 6 '15 at 12:07
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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.

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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.

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  • $\begingroup$ I understand this, that is why I answered the question in the first place because when I went through all the cases it was different to way the total number of ways should be $\endgroup$ Sep 6 '15 at 12:00

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