Open and Closed doors - amended

[Captain Spod]

If you have three doors that are locked COC, you can present that information in three ways: COC, CCO, and OCC. Four doors COCO can be presented six ways and five doors COCOC, eleven. What's the pattern; how can one predict how many ways there are of presenting the information for, say, twelve doors?

With apologies for earlier, inaccurate question.

[crofter]

Using C/O to denote CLOSED/OPEN Doors
and 5! to represent 5x4x3x2x1 = 120

COC has 3! / 2!1! = 3 "patterns"

COCO has 4! / 2!2! = 6 "patterns"

OCOCO has 5! / 2!3! = 10 "patterns"

2 C's and 10 O's have 12! / 2!10! = 66 "patterns"

[squarebear]

Surely 3 doors can also be OOO, CCC, OOC, OCO, COO making 7 options?

This is binary and the way to find out the possibilities for any number of doors is: 2^x-1.

4 doors will be (2^4)-1 or (2*2*2*2)-1 which equals 15.

Here are the 15 possibilities for 4 doors:
OOOO
OOOC
OOCO
OOCC
OCOO
OCOC
OCCO
COOO
COOC
COCO
COCC
CCOO
CCOC
CCCO
CCCC

12 doors would be (2^12)-1 which equals 4095 combinations.

[Rollo]

Assuming there is a requirement for the state of each successive door to be the opposite of the one before then there are two cases:-

Let the number of doors be n.

If n is odd there are (n-1)/2 open and (n+1)/2 closed

If n is even there are n/2 open and n/2 closed


For n odd the number of combinations is

n! / ((n-1)/2)! ((n+1)/2)!

For n even the number of combinations is

n! / (n/2)! (n/2)!

If n=3 then we have

3!/1!2! = 3

If n=4 then we have

4!/2!2! = 6

If n=5 then we have

5!/2!3! = 10 (not 11).


And if n=12 then we have

12! / 6!6! = 924

[Captain Spod]

Squarebear, Thanks for this but I did say the doors are locked. This is about presenting known information.

Crofter, Just what I wanted, thanks. Whats the ! called?

[Rose Maybud]

! is maths shorthand for factorial, so, for example,

factorial 4 is written 4! and means 4 x 3 x 2 x 1