Crosswords0 min ago
Open and Closed doors - amended
5 Answers
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.
With apologies for earlier, inaccurate question.
Answers
Best Answer
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For more on marking an answer as the "Best Answer", please visit our FAQ.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.
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.
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
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
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