Perms and Coms Dilemma (1 Viewer)

[Mathsisfun15]

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 behaves independently
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

 

[Drsoccerball]

I can explain why their method works. Each mouse has an option of 5 (5C1) and since theres 4 mouses: (5C1)(5C1)(5C1)(5C1)...

 

[Drsoccerball]

I can explain why their method works. Each mouse has an option of 5 (5C1) and since theres 4 mouses: (5C1)(5C1)(5C1)(5C1)...

and since the events are independent you can't calculate like you did (or at least thats what i think...)

 

[Mathsisfun15]

I worked out the error its in case 3 as it double counts

 

[Drsoccerball]

I worked out the error its in case 3 as it double counts

If something double counts it means your value is even lower now.

 

[Mathsisfun15]

If something double counts it means your value is even lower now.

I had a typo in the bottom line when I was making the sum. Through my case method it got 685 not the 585 that I originally typed