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The "combination" for a lock is actually a permutation, an ordered sequence. So for example: 1,2,3 and 3,2,1 are the same combination, but a different permutation. When opening a lock, obviously the sequence is important, so we want to calculate the permutations.
To do that, you multiply the number of possible choices for the first position times the number of possible choices for the second position, etc. Assuming that you can use the same number in all three positions (so 60 to the left, 60 to the right and 60 to the left is a valid choice), there are 60 x 60 x 60 possible permutations, or "combinations" for the lock = 216,000.
If you were not allowed to use the same number twice, it would be 60 x 59 x 58 = 205,320.
If you could repeat the same number, but not in sequence (so 20, 30, 20 is OK; but 20, 20, 30 is not), then there would be 60 x 59 x 59 = 208,860.
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How many combinations on a 3 number lock?
Just one. * * * * * Depends on how many numbers are on each ring. If there are x numbers, then the total number of combinations (actually they are permutations) is x*x*x or x3.
How do you reset i pod 4 digit passcode?
Go into settings, then general, then passcode lock, enter your passcode then click change passcode
What is graph based protocol in DBMS?
g ProtocolTree locking protocol is used to employ exclusive lock and when the database is in the form of a tree of data items. Tree locking protocol is serializable.Advantages of Tree Locking ProtocolUnlocking of data item is earlier.Shorter waiting time, increase in concurrency.Disadvantages of Tree Locking ProtocolTransaction may have to unnecessary lock data items to access its child nodes.Huge number of locks and locking overhead.
What Pierre traveled down a canal by barge from the town of Portage elevation 767 feet to the town of Barrage elevation 642 feet On the way he had to pass through a series of locks If each lock?
There were 5 of them.
How do you solve a permutation in math?
If there are n objects to fit r places (e.g. 9 people in 7 chairs, 4 tumblers in a lock) then the number of permutations is nCk, stated as n-choose-k. This number can be calculated by the formula n!/(n - k)!. If k is equal to n, then (n - k)! = 0! = 1, and the number of permutations is simply n!. If the direction of the permutation is irrelevant (e.g. ABCD is the same as DCBA) then divide by two to cancel out the double-counting.
