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Q: What are the possible six digit combinations you make out of the numbers 1 6?

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90

im not geeky enough to answer that

7

The number of possible 3 digit combinations you can make out of 1-9 with outrepeated digits is:9C3 = 9!/(3!(9-3)!) = 84

Not sure what a "didget" is. It is possible to make 18 5-digit numbers.

There are many combinations possible The most popular combination is 2,3,5,17,23.

There are 5*4*3*2 = 120 such numbers.

You can make 24 4-digit numbers. 1111 2222 3333 4444 5555 6666 1234 2345 3456 1356 6543 5432 4321 1432 ok i give up!

You can make 5 combinations of 1 number, 10 combinations of 2 numbers, 10 combinations of 3 numbers, 5 combinations of 4 numbers, and 1 combinations of 5 number. 31 in all.

0000, 0009, 0090, 0099, 0900, 0909, 0990, 0999, 9000, 9009, 9090, 9099, 9900, 9909, 9990, 9999

9999999999 100000000

Just one. In a combination, the order of the digits does not matter.

106 or a million.

There are many possible answers to this question ... for example, 9000, 2000, and 111. Can you be more specific?

4578, 4758, 5478, 5748, 5874, 5784, 7584, 7854, 7458, 7548, 8754, 8574

Simple enough to solve. The answer is a power of two. Assuming you have two possible digits, say for example, 3 and 4, then you simply have to multiply it by how many numbers you want to get the total number of combinations. Each number can be 3 or 4 in this case, and you have 5 numbers. That's two to the fifth. Five combinations of any two numbers. 2x2x2x2x2. The answer is 32 combinations.

You have 9 options for the first digit; whichever digit you choose you have 8 options for the second digit, 7 for the third digit, 6 for the fourth digit - so you have a total of 9 x 8 x 7 x 6 different combinations. Sure I could make a list, but that would be rather boring - and utterly useless as well.

From {1, 2, 3, 4} there are two prime numbers (2, 3} which can go in the hundreds position, which once chosen allows 3 possible choices for the tens digit and a further 2 possible choices for the units digit, giving 2 x 3 x 2 = 12 possible numbers.

There should be 125 combinations, but if you can only use the numbers once for each 3 digit number, then it should be 120. It's 125 because if the digits could be used more than once for each number then its the number on possible numbers (so there's 5 in this case, 2, 3, 5, 8, and 9) to the power of how big the digit is (in this case it's 3) It's 120 because if the digits could only be used once in each number then the combinations would be the number of possible numbers (again, 5) factorial, so 5!

A digit is that which we used to count numbers.

It depends upon whether the numbers can be used more than once. If the numbers can be used more than once, then there are 1,000 possible combinations; if not, then there are 720 possible combinations. ========== Assuming you are talking about integers you can calculate it this way: you can have any one of 9 digits for the first digit (zero is excluded because that would make it only a 2 digit number) You can have any one of 10 digits for the second and any one of 10 digits for the third digit. That gives you 9x10x10 = 900 different combinations for 3 digit numbers (not 1000). If you can include both negative and positive numbers as different numbers you get twice as many (2x900=18000). If you can count decimal numbers as 3 digit numbers (i.e. not restricted to integers) you can have 900 integers, 900 with the form xx.x, 1000 with the form x.xx (if zero is permitted to be the first digit and count as one of the 3 digits) or 900 (if a leading zero is NOT counted as one of the 3 digits). If a leading zero is NOT counted as one of the 3 digits, you could also have 1000 numbers of the form 0.xxx or just .xxx

This question does not make sense since a combination is independent of the order of its elements. That is, the combinations 1245 and 5142 are the same. Consequently, there can be no "last " numbers in a combination.

With 123 digits you can make 123 one-digit numbers.

9,000 - all the numbers between 1,000 and 9,999 inclusive. * * * * * NO. Those are PERMUTATIONS, not COMBINATIONS. Also, the question specified 4 digit combinations using 4 digits. The above answer uses 10 digits. If you start with 4 digits, you can make only 1 combination.

what is the greatest 4-digit number you can make with the digit 0,1,2,3, and 4