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There are 4 numbers between 0 and 3. Had this been a permutation problem, the answer would be 4!/2!=4*3=12, but this is a combinations problem. Since the duplicates (12 vs 21) come in pairs of 2, we divide the permutaions solution by 2!. Since 2! is just 2, 12/2 = 6 combinations in total.

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Vivian St. Cyr

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Q: How many 2 number combinations are there in the numbers 0 to 3?
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Related questions

How many combinations of 5 are there in 0 to 10 numbers?

There are a huge number of combinations of 5 numbers when using the numbers 0 through 10. There are 10 to the 5th power combinations of these numbers.


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How many combinations of 7 numbers are there using 0 through 9?

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