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Q: How many sets of 3-digit numbers can be made using 7 numbers and no duplicates within the 3-digit number?

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It depends on the set of numbers that you have and where, within that set, you start.

If you can repeat the numbers within the combination there are 10,000 different combinations. If you cannot repeat the numbers within the combination, there are 5040 different combinations.

Individual digits within a number have place values, not whole numbers.

The objects within a number set can be caled as "Elements" or "members".

11 is a prime and contains no composite numbers.

To make an even number with those numbers, it has to end in 6 because it's the only even number within the combination. The two possibilities are 576 or 756.

Within the real numbers, the logarithm of negative numbers is not defined.

The square of a "normal" number is not negative. Consequently, within real numbers, the square root of a negative number cannot exist. However, they do exist within complex numbers (which include real numbers)and, if you do study the theory of complex numbers you wil find that all the familiar properties are true.

Numbers within themselves can't be simplified. You have to have a certain number, a fraction for example, and then need to reduce it or change to a mixed number because of it being an improper fraction.

Yes, it is closed. This means that if you multiply two even number, you again get a number within the set of even numbers.

Within the real numbers, logarithms of zero and negative numbers are not defined. If you are interested in calculating logarithms within the complex number system, the Wikipedia article on "logarithm" has the necessary formulae. In this case, you can obtain the logarithm of a negative number, but not of zero.

Assuming all numbers are to be greater than zero and no duplicates exist within each possible answer, there are 5 possible combinations: 1 + 2 + 8, 1 + 3 + 7, 1 + 4 + 6, 2 + 3 + 6, and 2 + 4 + 5. If one number is allowed to be zero and no duplicates exist within each possible answer, there are 5 additional combinations: 1 + 2 + 8, 1 + 3 + 7, 1 + 4 + 6, 2 + 3 + 6, and 2 + 4 + 5. Removing the duplicates restriction allows: 0 + 0 + 11, 1 + 1 + 9, 2 + 2 + 7, 3 + 3 + 5, 4 + 4 + 3, and 5 + 5 + 1 Other solutions are possible if negative numbers are allowed.

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