What are all of the combinations to a 4-digit padlock?

Question

Can I get the exact and precise answer? For this could be the answer that many children and people in general have been curious, eager, and off-edge to discover. Thank you.

Answer 1

The answer is easy, it is of course 10,000. Every combination from 0000 to 9999 is possible (unless the hypothetical padlock has some restriction or limitations on it). Since this is the math of probability, the mathematical way to work it out would be:
possible outcomes for unit 1 * possible outcomes unit 2 * possible outcomes unit 3 * possible outcomes unit 4
10*10*10*10
=10,000

Answer 2

If you mean a padlock with 4 sets of numbers 0-9, then the answer is:

10*10*10*10 = 10,000 combinations

Answer 3

While i would agree with the above posters if the 4 digit padlock allowed repeated digits, in my experience, the padlock you so named does not allow repeated digits due to its mechanism.

so if suppose, one of the digits is 9, the other 3 digits cannot have 9 as a digit.

In this case the total numerical value of combinations is not 10x10x10x10=10000, but rather 10x9x8x7=5040

Answer 4

There are 10,000 possible combinations. Assuming that the dial is labeled from 0 to 9.

To see why this is the case, simply count from 0 up to 9999. Think of 0 as 0000, 1 as 0001, etc.

Answer 5

4 digits and each digit could be 0 - 9, so there are 10^4 = 10,000 possibilities.

If you are looking at one of those normal combination locks that go from 0 through 39 (the kind that you turn left then right then left) you have 40^3 combinations = 64,000 possibilities.

The number of possibilities = (the number of digits or numbers involved) ^ (the number of available numbers to use).

Answer 6

It is not that hard to figure out, just get a pencil and paper and start listing them 0000, 0001,0002,0003 ..... 9999, looks like there are ten thousand possibles.
Back in the day when personal computers came with BASIC interpreters, one could write a simple program to print them all;
FOR d1 = 0 TO 9
FOR d2 = 0 TO 9
FOR d3 = 0 TO 9
FOR d4 = 0 TO 9
PRINT d1; d2; d3; d4,
NEXT
NEXT
NEXT
NEXT
one might want to format the result a little better, and probably write to a file.

Answer 7

because of the fact that all known digits are used, there are ten thousand achieveable mixtures. Calculate it by utilizing taking the style of opportunities interior the 1st spot, 10, circumstances the quantity interior the 2d (10), the third, and fourth: 10*10*10*10 = ten thousand. attempting all the mixtures--merely count quantity from 0 to 9999--is somewhat easy to do with one among those lock. With in easy terms ten thousand mixtures, that is executed in some hours.

Answer 8

Well.. there are a total of 10,000 combinations, starting with 0000 and ending with 9999.

It would go like:

0000
0001
0002
0003
...
0499
0500
0501
...
0999
1000
1001
...
4999
5000
5001
...
9997
9998
9999

Tell me you can fill in the blanks yourself.

Answer 9

Unless I misunderstand your question it should just be from 0000 to 9999, making 10,000 combinations.

Answer 10

There are 10000 of them:

0000
0001
0002
0003
...

9999