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How many three digit combinations 000 999 are there without repeating the same number twice?
There are 10C3 = 10*9*8/(3*2*1) = 120 combinations.
Trying to figure something out. I have a three digit combination and it looks like this. 012345 012345 012345 I need all the possible three digit combinations like 000 or 001 or 345 like that.?
As you describe the problem, the condition is you can select any of the 6 numbers with replacement since you have 3 sets of numbers for each value. Per the related link, you have N = 6 and n = 3. The number of combinations is Nn, or 63, or 216. If it is easier to think of the problem in this way, here is another way to look at it. You have 6 choices for the first digit, 6 choices for the second digit, and 6 choices for the third digit. So, you have 6*6*6 = 216 combinations.
How many 10-digit even numbers can be formed if the digits can be repeated?
Assuming that leading zeros are not permitted (that is the lowest ten digit number is 1 000 000 000), then there are 4 500 000 000 possible even numbers
How many three digit combinations can be made from numbers zero through nine?
If order is important and repetition is allowed, then : 000 through 999 (1000 possible). If no leading zeros, then start at 100 through 999 (900 possible). See related link, if you need to limit to no repetition, or order doesn't matter (like if 123 is the same as 231).
How many 3 digit even numbers are there?
There are 450 even three-digit integers, from 100 to 998. This excludes numbers with leading zeroes (000 to 099).
How many 3 digit combinations are there in the numbers 1 to 9?
It depends. If you can only use each number once, there are 720 combinations. If you can use numbers multiple times, then there are 1000 combinations, by using all numbers from 000 to 999.
If using numbers 0 through 9 how many 3 digit combinations?
1,000. The list looks just like the counting numbers from 000 to 999 .
How many combinations are there from 000 to 999?
1000
How many 3 digit identification tags can be made the digits can be used more once?
1000: from 000 to 9991000: from 000 to 9991000: from 000 to 9991000: from 000 to 999
How many three digit combinations 000 999 are there without repeating the same number twice?
There are 10C3 = 10*9*8/(3*2*1) = 120 combinations.
Trying to figure something out. I have a three digit combination and it looks like this. 012345 012345 012345 I need all the possible three digit combinations like 000 or 001 or 345 like that.?
As you describe the problem, the condition is you can select any of the 6 numbers with replacement since you have 3 sets of numbers for each value. Per the related link, you have N = 6 and n = 3. The number of combinations is Nn, or 63, or 216. If it is easier to think of the problem in this way, here is another way to look at it. You have 6 choices for the first digit, 6 choices for the second digit, and 6 choices for the third digit. So, you have 6*6*6 = 216 combinations.
How many 10-digit even numbers can be formed if the digits can be repeated?
Assuming that leading zeros are not permitted (that is the lowest ten digit number is 1 000 000 000), then there are 4 500 000 000 possible even numbers
What are the possible combinations for a lock with 3 wheels and the digits 0 to 9?
The short answer is 1000. This is very easy to visualise: Simply consider each number in the combination to be a digit in a decimal number. We then end up with a three-digit number. Such a three-digit number ranges in value from 000 to 999, or 1000 unique combinations.
How many three digit combinations 000-999are there without repeating the same number twice?
999
How many number combination can made from four numbers?
10 000 * * * * * NO! That is the number of PERMUTATIONS, not COMBINATIONS. In a combination, the order does not matter so that 1234 is the same as 1432 or 3412 etc. Assuming the 4 numbers are different, the correct answer is 15 comprising 4 1-digit combinations, 6 2-digit combinations, 4 3-digit combinations and 1 4-digit combination. Another way to look at it is that the first number can be in a combination or not. With each of these possibilities, the second can be in or out - giving 2*2 = 4 ways so far. With each of these there are two options for the third giving 2*2*2 = 8 combinations so far and then the last number makes it 2*2*2*2 = 16. But one of these combinations contains none of the numbers - each one is not in. Leaving that one out gives the answer 15. In general, the number of combinations of any size, from n distinct objects is 2n and if you exclude the null combination, it is 2n - 1.
How many six digit numbers can be formed using the digits 0123456789 Example equals 123456 or 098765?
The number of six digit numbers that you can make from ten different digits ifrepetitions of same digit on the six digit number is allowed is 1 000 000 numbers(including number 000 000).If no repetitions of the the same digit are allowed then you have:10P6 = 10!/(10-6)! = 151 200 different six digit numbers(six digit permutations form 10 different digits).
How many 10 digit phone numbers are possible?
Without restrictions, it was would numbers 000-000-0000 through 999-999-9999. So that would be 9,999,999,999 + 1 = 10 billion different 10-digit phone numbers. Ex: If there existed single digit phone numbers, there would be 10, because the digits are 0 through 9. If there existed only double digit phone numbers, then it would be 00 through 99 which would be 100 total two-digit numbers. Therefore the total possible combinations for an X digit phone number would be: 10^X
