praveen_gmat wrote:kmittal82 wrote:At least 8 distinct digit means at least 8! possible passwords
12 seconds to try 1 password, hence minimum time needed = 8! x 12, whats the OA?
I think the question here is trying to trick you by using the word "combination", whereas this is really a permutation problem, since the order matters (strictly speaking all "combination locks" should be called "permutation locks"

)
What do you mean by OA?
The answer you have given is wrong. The answer is (
10P8 + 10P9 + 10P10) × 12 seconds, though I dint understand how it is that way ..
On the actual GMAT, the answer will never say "10P8" - what's the source of this question? It's always beneficial to provide the answers choices and the source, so readers know what alternative solutions are available and whether the source is reliable.
"10P8" refers to the permutations formula:
nPk = n!/(n-k)!
in which n is the total number of objects available and k is the number of objects that you're actually arranging.
So:
for a code with 8 distinct digits, we have 10 digits available (0, 1, 2, ..., 9) and are using 8 of them;
for a code with 9 distinct digits, we have 10 digits available and are using 9 of them; and
for a code with 10 distinct digits we have 10 digits available and are using all 10 of them.
Since we want an 8 digit OR a 9 digit OR a 10 digit code, we add the individual results to get the total number of codes.
10P8 + 10P9 + 10P10
= 10!/2! + 10!/1! + 10!/0!
= 10!/2 + 10! + 10!
(0! = 1)
and, since it takes 12 seconds per try, to
guarantee that we hit the code we multiply by 12 seconds:
12(10!/2 + 10! + 10!)