Question 1183743
there is only one possible code.
the possible digits are 2,4,6,8.
can't be 0
can't be any odd number between 0 and 9.
the probability that the first digit is the right digit is 1/4.
the probability that the second digit is the right digit, assuming that the the first digit is correct, is 1/3.
the probability that the third digit is the right digit, assuming that the first two digits are correct, is 1/2.
the total probability is 1/4 * 1/3 * 1/2 = 1/24.


if you look at the number of permutations possible, you will get 4P3 = 4! / (1!) = 4*3*2*1 / 1 = 24.


only one of those permutations is the correct permutation, therefore the probability is 1/24.


the permutation formula is p(n,x) = n! / (n-x)!.
when n = 4 and x = 3, the formula becomes p(4,3) = 4! / (4-3)! = 4! / 1! = 24.


only one permutation out of those 24 permutations is the correct permutation,  therefore the probability is 1/24.


i'm pretty sure this is correct.
i was able to get the same answer both ways.
the denominator is the permutation of 4 possible numbers taken 3 at a time.
the numerator is 1.
the possible number are 2,4,6,8.


the possible 3 digit permutations are:


2 4 6 *****
2 6 4
4 2 6
4 6 2
6 2 4
6 4 2


2 4 8 *****
2 8 4
4 2 8 
4 8 2
8 2 4
8 4 2


2 6 8 *****
2 8 6
6 2 8
6 8 2
8 2 6
8 6 2


4 6 8 *****
4 8 6
6 4 8
6 8 4
8 4 6
8 6 4


note that you have 4 sets of 6 = 24.


each set of 6 is a separate combination, if you don't take order of the digits into account


the permutation formula is p(n,x) = n! / (n-x)!


the combination formula is c(n,x) = n! / (x! * (n-x)!)


when n = 4 and x = 3, these formulas become:


p(4,3) = 4! / 3! = 4! = 24.


c(4,3) = 4! / (3! * 1!) = 4! / 3! = 4.


you have 4 possible combinatins where order is not important.
they are:


2 4 6
2 4 8
2 6 8
4 6 8


within each of those you have 6 possible arrangements where order is important.


4 * 6 = 24.


he will need to try up to 24 possible permutations and is assured to get the right combination.