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How many different 4-digit codes are possible if the digits cannot be repeated?
If you include 0000, ten thousand unique four digit codes are possible.
How many different 4-digit codes are possible if the last digit has to be zero?
There can be 103 = 1000 codes.
How many threedigit numbers can you make if the first numbercan be 1234 the second number can be 123456 and the third 012?
The first digit has 4 choices for its digit. The second digit has 6 choices and the third has 3. The solution would simply be 4*6*3=72 three digit numbers.
How many 5 digit zip codes are possible?
yes of course I know people with 3 and 2 digit addresses
How many possible 4 digit 0 thru 9 codes?
If the first digit can be zero, there are 5,040. If the first digit can't be zero, there are only 4,536.
How many different 4-digit codes are possible if the digits cannot be repeated?
If you include 0000, ten thousand unique four digit codes are possible.
How many different 4-digit codes are possible if the last digit has to be zero?
There can be 103 = 1000 codes.
How many 5 digit zip codes are possible if digits cannot be repeated?
100,000 - Every number from 00000 to 99999.
How many possible 4 digit combinations are there using 0-9. how many different combination codes are possible?
There are 210 4 digit combinations and 5040 different 4 digit codes.
How many area codes can you make without 0 and 1 in the first digit and the second has to be 0 and 1?
With the restrictions specified in your question, there are 8 x 2 x 10 = 160 possible area codes. However, those restrictions do not match the restrictions on actual area codes in North America (USA, Canada, etc.). North American area codes have the first digit not 0 or 1, the second digit not 9, and the third digit not the same as the second digit (such codes are reserved for special purposes like toll-free), leaving 8 x 9 x 9 = 648 possible area codes.
How many threedigit numbers can you make if the first numbercan be 1234 the second number can be 123456 and the third 012?
The first digit has 4 choices for its digit. The second digit has 6 choices and the third has 3. The solution would simply be 4*6*3=72 three digit numbers.
How many three digit even numbers are possible if the left digit cannot be zero?
450
How many possible 3 digit area codes can be created without repeating a number?
The number of 3-digit numbers with no repeated digits is simply 10x9x8 = 720, if you allow, for example, 012 as a 3-digit number. There are 10 digits, any of which might be the first digit. The second digit can be any digit except the digit that was used for the first digit, leaving 9 possibilities. The third digit then has 8 possibilities, since it can't be the same as the first or second digit. The actual number of possible area codes will be lower, because there are additional restrictions on the number combinations for a valid area code. For example, in North America (USA, Canada, etc.), the first digit of an area code cannot be 0 or 1 and the middle digit cannot be 9.
How many 5 digit zip codes are possible?
yes of course I know people with 3 and 2 digit addresses
How many possible 4 digit 0 thru 9 codes?
If the first digit can be zero, there are 5,040. If the first digit can't be zero, there are only 4,536.
How many codes are possible with 2 letters and one digit if no letter or digit appear more than once?
There are 26 choices for the first letter, 25 choices for the second letter (since we can't repeat the first letter), and 10 choices for the digit. Therefore, the total number of possible codes is: 26 x 25 x 10 = 6,500 So there are 6,500 possible codes with 2 letters and one digit if no letter or digit appears more than once.
How many four digits numbers can be formed using 0123456789?
Assuming that the first digit of the 4 digit number cannot be 0, then there are 9 possible digits for the first of the four. Also assuming that each digit does not need to be unique, then the next three digits of the four can have 10 possible for each. That results in 9x10x10x10 = 9000 possible 4 digit numbers. If, however, you can not use the same number twice in completing the 4 digit number, and the first digit cannot be 0, then the result is 9x9x8x7 = 4536 possible 4 digit numbers. If the 4 digit number can start with 0, then there are 10,000 possible 4 digit numbers. If the 4 digit number can start with 0, and you cannot use any number twice, then the result is 10x9x8x7 = 5040 possilbe 4 digit numbers.
