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Each digit can appear in each of the 4 positions.

There are 9 digits, therefore there are 9⁴ = 6561 such combinations.

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Can you list the 6561 combinations?

Q: If using numbers 1 through 9 how many different 4 digit combinations can be made with no restrictions on order or repeats?

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Assuming you are using combinations in the colloquial way (which is the mathematical "permutations" where order of selection does matter) to create a 3 digit number that does not start with 0, ie creating a number that is between 100 and 999 inclusive then: If repeats are not allowed there are 3 × 3 × 2 = 18 possible numbers If repeats are allowed, then there are 3 × 4 × 4 = 48 possible numbers. If you are using combinations in the mathematical sense where order of selection does not matter and are creating groups of 3 digits, then: If repeats are not allowed there are 4 possible groups If repeats are allowed there are 20 possible groups.

you could make a probability tree if you could be bothered

The number of combinations is 50C6 = 50*49*48*47*46*45/(6*5*4*3*2*1) = 15,890,700

Without repeats there are 4 × 3 = 12 possible 2 digit numbers. With repeats there are 4 × 4 = 16 possible 2 digit numbers.

there are 10 numbers that you can choose from 0-9 you are choosing 8 at a time, meaning no repeats so the answer is "10 choose 8" or 10!/(8!*(8-2)!)

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In other words, how many 4 digit combination locks are there using the digits 0-9 on each wheel. There are 10×10×10×10 = 10⁴ = 10,000 such combinations.

Without repeats, 24. With repeats, 256.

If repeats are allowed than an infinite number of combinations is possible.

Assuming you are using combinations in the colloquial way (which is the mathematical "permutations" where order of selection does matter) to create a 3 digit number that does not start with 0, ie creating a number that is between 100 and 999 inclusive then: If repeats are not allowed there are 3 × 3 × 2 = 18 possible numbers If repeats are allowed, then there are 3 × 4 × 4 = 48 possible numbers. If you are using combinations in the mathematical sense where order of selection does not matter and are creating groups of 3 digits, then: If repeats are not allowed there are 4 possible groups If repeats are allowed there are 20 possible groups.

Irrational numbers can not be repeating decimals. Any number that is a repeating decimal is rational.

you could make a probability tree if you could be bothered

The number of combinations is 50C6 = 50*49*48*47*46*45/(6*5*4*3*2*1) = 15,890,700

Without repeats there are 4 × 3 = 12 possible 2 digit numbers. With repeats there are 4 × 4 = 16 possible 2 digit numbers.

there are 10 numbers that you can choose from 0-9 you are choosing 8 at a time, meaning no repeats so the answer is "10 choose 8" or 10!/(8!*(8-2)!)

We are not told anything about repeats so I will assume you can repeat a digit. For example, you can have 111111 The first digits has 50 choices so does the second and the third..etc so we have 50^6 combinations.

Assuming leading zeros are not permitted, then: If repeats are not allowed there are 30 possible numbers. If repeats are allowed there are 60 possible numbers.

If repeats are permitted: 2 x 5 x 5 = 50 different odd numbers If repeats are not permitted: 2 x 4 x 3 = 24 different odd numbers