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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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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.
How many number combinations are possible with three numbers?
If the numbers can be repeated and the numbers are 0-9 then there are 1000 different combinations.
How many 4-number combinations using numbers 0-9?
9000
How many 4 digit combinations can be made from the numbers 0 to 9?
If every number can be used as many times as you like, there are 104 = 10000 different combinations. If each number can only be used once, there are 9!/(9 - 4)! = 5040 combinations.
How many 4 number combinations are there in the numbers 0 to 7?
There are 10,000 four number combinations in the numbers 0 through 9. * * * * * No, that is simply not true. The combination 1234 is the same as the combination 2134 and 2314 and 1423 etc. So, there are just 8*7*6*5/(4*3*2*1) = 70 combinations.
