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Assuming you're allowed to repeat numbers, the answer is 81. For each digit, you have three choices: 2, 5, and 6. So there are three one-digit numbers you can make, and each one digit number can be combined with any of three other digits to make a two-digit number, giving 9 (3 times 3) possibilities. Now you can take any of the nine two-digit numbers as the tens and ones places in the four-digit number and any other two-digit number as the thousands and hundreds places, which gives 81 (9 times 9) possibilities.
If you can't repeat numbers, the answer is zero because you don't have enough numbers to fill all the digits.
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How many 4 digit combinations can be made from 10 numbers with out repeating?
The answer is 10C4 = 10!/[4!*6!] = 210
If using numbers 0 through 9 how many different 2 digit combinations can be made?
If there are no restrictions on duplicated numbers or other patterns of numbers then there are 10 ways of selecting the first digit and also 10 ways of selecting the second digit. The number of combinations is therefore 10 x 10 = 100.
How many combinations of three one digit numbers equal 4?
5
How many combinations of 2 digit numbers can be produced from 01 99?
45.
If using numbers 1 through 9 how many different 4 digit combinations can be made with no restrictions on order or repeats?
Each digit can appear in each of the 4 positions. There are 9 digits, therefore there are 9⁴ = 6561 such combinations.
