Question 1111291: When testing for current in a cable with eight color-coded wires, the author used a meter to test three wires at a time. How many different tests are required for every possible pairing of three wires?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! you would use the combination formula of:
c(n,x) = n! / (x! * (n-x)!)
n = 8
x = 3
formula becomes c(8,3) = 8! / (3! * 5!)
result is c(8,3) = 56 different test would be required to get all possible combinations of 3 wires each, where order within each set of 3 is not important.
to see how this works, make n smaller so the number of possible combinations will be less.
when n = 4, c(4,3) = 4
let the wires be a, b, c, and d.
the test required to make sure you got all possible combinations would be:
abc
abd
acd
bcd
note that the combination formula assumes order is not important within each set of 3.
if order is important, then the permutation formula would be used.
that formula is p(n,x) = n! / (n-x)!
note that the x! in the denominator is missing in the permutation formula, while it is present in the combination formula.
the x! in the denominator is the one that takes away the order is important part.
that's what turns the permutation formula into the combination formula.
the number of possible permutations with p(4,3) is 4! / (4 - 3)! = 4! / 1! = 24
those permutations would be:
abc ***** original
acb
bac
bca
cab
cba
abd ***** original
adb
bad
bda
dab
dba
acd ***** original
adc
cad
cda
dac
dca
bcd ***** original
bdc
cbd
cdb
dbc
dcb
out of the 24 possible permutations, there are only 4 possible combinations.
note that, for each of the originals that were derived from the combination formula, there are 6 possible permutations.
3! = 6, which is where the x! in the denominator comes from in the combination formula.
c(n,x) = n! / (x! * (n-x)!) becomes c(4,3) = 4! / (3! * (4-3)!) which becomes c(4,3) = 4! / (3! * 1!) = 4
p(n,x) = n! / (n-x)! becomes p(4,3) = 4! / (4-3)! which becomes p(4,3) = 4! / 1! = 24
bottom line:
you have 8 wires and you want to test all possible combinations of 3 of these at a time.
order, or arrangement of the wires within each set of 3 is not important, therefore you use the combination formula and not the permutations formula.
c(8,3) = 8! / (3! * 5!) = 56 possible sets of 3 wires where order within each set of 3 is not important.
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