Help with 4 digit combination lock?

Question

I have a four digit combination lock I forgot the code to. It is four digits with no repating digits, ie. 1123, 4565, and 9878 are invalid. How many possiable codes are there is there an easy way to generate a list?

Answer 1

1*2=2*3=6*4=24*5=120*6=720*7=5040*8=40320*9=362880

362,880 possible combinations or 9!

Answer 2

The number of combinations with 10 different, non repeating numbers is 5040. This is found using the concept of factorials.
10!/6!= (10*9*8...*2*1)/(6*5*4...*2*1)= 10*9*8*7= 5040.

If you could have a number repeat itself, the answer would have been 10*10*10*10= 10,000 which is verified since the numbers range from 0000 to 9999.

Answer 3

There are 10 options for the first number, then, since that number cannot be repeated, there are 9 options for the second number, then 8, then 7.
Thus (10x9x8x7) = 5040 different options.

I don't believe I could give you an easy way to generate a list...I can't think of an easy way to generate a list of any 5040 numbers. I would write a little macro in Excel, I guess.

Answer 4

You'll almost certainly not figure it out on your own if you didn't set the code initially. Your best bet is to contact the manufacturer of the lock, give them the serial number, and ask them for the combination.

Answer 5

easy way to generate this list:

if you know C or some programming language, the following code should do it:

int loop1, loop2, loop3, loop4;

for (loop1 = 0; loop1 < 10; loop1++) {
for (loop2 = 0; loop2 < 10; loop2++) {
if (loop1 == loop2) continue;
for (loop 3 = 0; loop3 < 10; loop3++) {
if (loop1 == loop3) continue;
if (loop2 == loop3) continue;
for (loop4 = 0; loop3 < 10; loop4++) {
if (loop1 == loop4) continue;
if (loop2 == loop4) continue;
if (loop3 == loop4) continue;
printf ("%d%d%d%d\n",loop1,loop2,loop3,loop4);
}
}
}
}

Just a few notes:
continue is used to jump out of the loop at the current iteration and go to nested loop. It will increase the loop, so for the following example:

loop1 = 2;
loop2 = 3;
loop3 = 3;
loop2 == loop3 is true, so don't even go to loop4

next loop iteration would look like:

loop1 = 2;
loop2 = 3
loop3 = 4;
///start going through loop4

Answer 6

make clcultion binary