how many possible combinations in a 4 digit code?
10234
There are 10,000 possible combinations that the digits 0-9 can be arranged into to form a four-digit code.
814112
SOLUTION: how many 4 digit number combinations can u make out of this 4 numbers 1486 and what are they. ANS. IS 24.
22327
There are 1 billion 9 digit numbers (000,000,000 through 999,999,999). There are 45 different combinations of two different numerals (10 x 9 divided by 2). There are 512 (2 to the 9th power) different permutations for any two numbers to be used in a 9 digit number.
600213
There are 10,000 possible combinations that the digits 0-9 can be arranged to form a 4-digit pin code.
643296
Explanation: Since there are no parameters as to use of numerals we have nine numbers that can be inserted into each of four digits of the code. This means there are 6561 possible combinations for the key code.
77076
Assuming no five-digit number can begin with zero, there are 9 possible choices for the first digit. Then there are 10 possible choices for each of the remaining four digits. Therefore, you have 9 x 10 x 10 x 10 x 10 combinations, or 9 x 10^4, which is 90,000 different combinations.
535085
['Answer: There are 9,999 different combinations in a 4-digit lock.', 'Explanation: The 4-digit number on a lock begins from 0000 and ends up to 9999 which are:', 'From 0000 to 0999 there are 999 combinations. ( ... ', 'From 1,000 to 9,999 there are 8,999 combinations (9,000)', 'And now we have 1,000+9,000=10,000 combinations total.']
431600
We conclude that there are 35−3×25+3=150 combinations. If there were no restrictions, the number of five digit codes that could be formed with the digits 1, 2, 3 would be 35 since there are three ways to choose each of the five digits.
527447
2! See for 11 numbers there can be 9 ways for each number for 1st nine places since 0 can't be the fist number and next two are repeated from 10 numbers hence 10!/2! So 99 combinations have 10!/2! Permutations hence total ways are 99*10!/2!
397405
Could it be 10C3? The safe combination could be 433 or 334; the combinations are the same, but their ordering is different. Order is important for the safe combinations, this is a permutation problem. A safe code can be made of any of the numbers {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.
966160
Hence, in this case there are 4 3 = 12 possible two digit numbers. For the third digit there are only 2 choices and for the last digit, one choice. Thus if you are not allowed to repeat a digit the number of possible 4 digit numbers you can construct from 1,2,3,4 is 4 3 2 1 = 24.
1212839
24. Well you have n choices for the nth digit. In your case you have 4 digits. So 4x3x2x1=24 combinations of all 4 digits with none repeating.
820814
So there are 4 x 3 x 2 x 1 = 24 possible ways of arranging 4 items. Therefore I divide 5040 / 24 = 210. So there are 210 different combinations of four digits chosen from 0-9 where the digits don't repeat.
768495
The number of combinations is 8,145,060. This is because there are 45!/(6!
419263
A standard 3-digit lock has 999 possible combinations, which is why the 4- digit locks cost a bit of a step more money, as they have 9999 possibles. One very important thing to remember is the resetting of this lock to something much different from your actual combo, when you lock it.
1462999
I believe there are 126 combinations with 4 dice. This is equal to (6x7x8x9)/(1x2x3x4). Pattern confirmed.
96505
Hence there are 2 choices for the fourth digit and only 1 choice for the fifth digit. Thus you have made 5 × 4 × 3 × 2 1 = 120 choices and there are 120 possible 5 digit numbers made from 1, 2, 3, 4 and 5 if you don't allow any digit to be repeated. Now consider the possibilities with 13 as the first two digits.