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There are a total of 1,000 three-digit combinations from 000 to 999. This includes all combinations where the digits can range from 0 to 9, allowing for repetitions. Each of the three digit positions can have 10 possible values (0-9), leading to (10 \times 10 \times 10 = 1,000) combinations.
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How many three digit combinations can be made?
To find the number of three-digit combinations, we consider the digits from 000 to 999. Each digit can range from 0 to 9, giving us 10 options for each of the three digits. Therefore, the total number of three-digit combinations is (10 \times 10 \times 10 = 1,000).
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).
