My 627-3 case has 357 for the left hand combo and 753 for the right hand combo.
Now we all know your secret....You'll have to change them now!!!!!!
JIM...........
My 627-3 case has 357 for the left hand combo and 753 for the right hand combo.
I'd agree with getting a locksmith and trying to get the seller to pay but expect he'd balk too.
I have one of those fireproof document cases that is locked and I think has papers from my late Father's estate. I can't find a key to fit and will try a locksmith too. I don't want to cut the locks as I'd like to reuse the case.
I've been to a couple of Michigan gun shows this year where a vendor was selling lock pick kits. PM me if you want his website info.
If it's a $20 fire box, pry it open and then buy a new one.
Doesn't sound like it, does it?Bought a 627, complete with performance center aluminum locking case.
Doesn't sound like it, does it?
PS Once you figure it out, write it down some place where you will not forget where you put it, but don't tape it to the case. LOL
Jus' Thinkin'
If order mattered, we would say select the first number (10 choices), then the second (9 choices) then the third (8 choices). So there would be 10 x 9 x 8 = 720 possible choices.
But because order does not matter, we have to take care of duplicates as you mentioned. How many duplicates are there for each set of three numbers? Well, again, we can choose one of the three as the "first", so there are 3 choices for that, then 2 choices for the "second" digit, then 1 choice for the last digit. There are, you see, 3 x 2 x 1 = 6 possible ways of arranging the three digits.
Therefore in that set of 720 possibilities, each unique combination of three digits is represented 6 times. So we just divide by 6.
720 / 6 = 120.That's your answer.
These are what are mathematically called "combinations". You can use a formula involving factorials to determine the number of combinations.
In this case, we say this is "10 Choose 3" and write it as 10C3. That means from a set of ten (digits in
this case), choose 3 regardless of order.
The formula for nCm is
nCm = n! / [ m! x (n-m)!]So in your question, we have
10C3 = 10! / [3! x (10-3)!] = 10! / [3! 7!]
= (10 x 9 x 8 x 7!) / [ (3 x 2 x 1) 7!]
= (720) / 6 ... because the 7! on the top and bottom cancels.
= 120 as we expected.