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Q: How many different four digit combinations are there of the numbers zero through nine and using one number twice in the combination?

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There is only one combination since the order of the numbers in a combination does not matter.

10!/3! = 604800 different combinations.

Assuming that repeated numbers are allowed, the number of possible combinations is given by 40 * 40 * 40 = 64000.If repeated numbers are not allowed, the number of possible combinations is given by 40 * 39 * 38 = 59280.

Exactly 3,628,800, or 10!.

1 and 11 and 21 and 31 and 42 and 12 and 22 and 32 and 43 and 13 and 23 and 33 and 44 and 14 and 24 and 34 and 416 combinations

Related questions

How many numbers per combination?

There is only one combination. In a combination the order of the numbers does not matter so the only combination is 0123456789. This is the same as 1326458097

9^9 = 387420489 different digit combinations.

There are 3,628,800 different permutations that can be made when using the numbers 0 through 9, if each number is only used once in each combination.

Including the null combination, there are 2^14 = 16384 combinations.

There is 1 combination of all ten numbers, 10 combinations of one number and of nine numbers, 45 combinations of two or eight numbers, 120 combinations of three or seven numbers, 210 combinations of four or six numbers and 252 combinations of five numbers. That is 1023 = 210 - 1 in total.

There is only one combination since the order of the numbers in a combination does not matter.

1000

10!/3! = 604800 different combinations.

66

15

2304

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