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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.

Q: How many 4 digit combinations can be made from these 3 numbers 2-6-5?

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There are 38760 combinations.

The answer is 10C4 = 10!/[4!*6!] = 210

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.

5

45.

Related questions

Number of 7 digit combinations out of the 10 one-digit numbers = 120.

The answer will depend on how many digits there are in each of the 30 numbers. If the 30 numbers are all 6-digit numbers then the answer is NONE! If the 30 numbers are the first 30 counting numbers then there are 126 combinations of five 1-digit numbers, 1764 combinations of three 1-digit numbers and one 2-digit number, and 1710 combinations of one 1-digit number and two 2-digit numbers. That makes a total of 3600 5-digit combinations.

There are 38760 combinations.

120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.

It is: 9C7 = 36

66

15

6 ways: 931,913,139,193,391,319

10,000

10 Combinations (if order doesn't matter). 3,628,800 Possiblilities (if order matters).

There are 1140 five digit combinations between numbers 1 and 20.

The answer is 10C4 = 10!/[4!*6!] = 210