Different permutation and combination

lingping7

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Jan 6, 2013
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The question :
There is a 4 digit bike lock and each digit can be from 0-9. The correct code was forgotten, but the person knew that the lock had EXACTLY one 3
What is the maximum number of combinations that have to be tried to open the lock.


What I did was that suppose the unit digit contains 3, then for the other numbers, there are 1000 possible combinations - 000, 001, 002 till 997, 998, 999

This way by keeping 3 in each place value, it turns out to be 4000 combinations.

But I made a made a mistake not reading exactly ( reason in caps)

So if there is only one 3 in the combination, how do I know the maximum combinations need to be tried to open the lock

Thanks in advance.
 
The question :
There is a 4 digit bike lock and each digit can be from 0-9. The correct code was forgotten, but the person knew that the lock had EXACTLY one 3
What is the maximum number of combinations that have to be tried to open the lock.


What I did was that suppose the unit digit contains 3, then for the other numbers, there are 1000 possible combinations - 000, 001, 002 till 997, 998, 999

This way by keeping 3 in each place value, it turns out to be 4000 combinations.

But I made a made a mistake not reading exactly ( reason in caps)

So if there is only one 3 in the combination, how do I know the maximum combinations need to be tried to open the lock

Thanks in advance.
It is a 4-digit code, so all numbers from 0000 to 9999 have to be considered. Now each digit either IS a 3 (probability p=1/10), or it is NOT a 3 (probability q=9/10). How many have the first digit a 3, and no others: That is, 3xxx where x is not 3:
N(yes no no no) = 1 * 9^3
Multiply by 4 because the probability on any digit position is the same, x3xx, etc.
 
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