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Q: If using numbers 0 through 9 how many different 2 digit combinations can be made?

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10!/3! = 604800 different combinations.

9!/6!, if the six different orders of any 3 digits are considered distinct combinations.

There are 360 of them.

9

Exactly 3,628,800, or 10!.

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

9^9 = 387420489 different digit combinations.

66

15

10,000

2304

56 combinations. :)

10!/3! = 604800 different combinations.

Through the magic of perms and coms the answer is 729

9!/6!, if the six different orders of any 3 digits are considered distinct combinations.

This question needs clarificatioh. There are 4 one digit number combinations, 16 two digit combinations, ... 4 raised to the n power for n digit combinations.

There are 360 of them.

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