2.5L R3 pannier problem

[Jay]

are you holding the lock upside down? Otherwise what @technoguppy suggested. That has happened to me with setting dial locks.

 

[Ygransom]

There is a combination lock on the equipment locker at work that is very difficult to open as the numbers must be offset from the mark. IF you put center of number on mark it won't open, but if you put bottom of numbers on mark it opens smoothly. Might try offsetting numbers when you put them in

 

[Tripps]

I had a safe that apparently wore with age, a couple of the numbers drifted one digit, but that was a dial combination

 

[Mark Olson]

I had a safe that apparently wore with age, a couple of the numbers drifted one digit, but that was a dial combination

I've got a good dealer!! He's going to replace the bag and return my faulty one under warranty!! Now I just need to figure out how to get my sheep skin seat cover out of it Thanks for all the ideas, I tried all of them and a few more. It initially took my new code just like the other one did. Then a day later it just flat would not open. Who knows, malfunction I guess.

 

[1olbull]

Well, only 999 numbers to try. Oops, 1000, forgot 000

No so fast there, Amigo!
Three digit combos using 0 - 9 are 720 possibilities.

 

[ruzzle]

That's if order matters. Which in this case it probably doesn't?

If it doesn't (e.g. 123 and 321) then how many?

 

[MrPix]

No so fast there, Amigo!
Three digit combos using 0 - 9 are 720 possibilities.

Err... OK. 3 columns, 0-9 = 10 numbers to the power of 3.... 10 * 10 * 10 = 100 * 10 = 1000.... just saying

 

[1olbull]

Err... OK. 3 columns, 0-9 = 10 numbers to the power of 3.... 10 * 10 * 10 = 100 * 10 = 1000.... just saying

If what you want are all possible three digit numbers with no repetition of the digits then you have 10 choices for the first digit, you have 9 choices for the 2nd digit, and you have 8 choices for the 3rd digit giving you 10x9x8 = 720 in all.

 

[MrPix]

If what you want are all possible three digit numbers with no repetition of the digits then you have 10 choices for the first digit, you have 9 choices for the 2nd digit, and you have 8 choices for the 3rd digit giving you 10x9x8 = 720 in all.

I can't think of a single repeated number between 000 and 999 ... can you give me a single example of a repeated sequence out of the 280 you have calculated that are there please?

 

[technoguppy]

As you cannot repeat the numbers it is a permutation (not a combination) for the sake of calculating the number of possible outcomes.

The total number is 10! / (10 - 3)!, in English Ten Factorial divided by the Factorial of ten (number of available digits) minus three (number of times the digit is used).

or 10! / 7! = 3,628,800 / 5,040 = 720