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Q: What are the possible 4 digit combinations 0-9 with 1 as the third number?

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there are 10 possibilities for the first spot, 9 for the second, 8 for the third 10x9x8=720 combinations

5,040Assuming that the combination uses ALL single digits from 0 to 9, then for the first digit you will be able to use all 10 numbers, for the second digit you will be able to use 9, for the third digit 8 and for the last digit 7, giving a total number of combinations of 10 x 9 x 8 x 7 = 5,040 without the same number being used more than once in each combination.

Assuming that the digits 0-9 are available, the answer is 5040. This is because for the first digit you can pick any of the original 10 digits. For the second you can have any otherdigit, so 9 possible choices. The third follows the same rule, so has 8 possibilities, and the forth digit, 7.This means that the total number of combinations is 10 x 9 x 8 x 7 = 5040

If the order of the numbers are important, then this is a simple combination problem. There are 10 possible numbers to choose from for the first number. Then there are 9 options for the second number. Then there are 8 options for the third, and so on. Thus, the number of possible combinations can be calculated as 10x9x8x7x6x5. This comes out at 151,200 possible 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.

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there are 10 possibilities for the first spot, 9 for the second, 8 for the third 10x9x8=720 combinations

5,040Assuming that the combination uses ALL single digits from 0 to 9, then for the first digit you will be able to use all 10 numbers, for the second digit you will be able to use 9, for the third digit 8 and for the last digit 7, giving a total number of combinations of 10 x 9 x 8 x 7 = 5,040 without the same number being used more than once in each combination.

You have a three digit number that can only use each digit once in combination. Therefore, a number like 456 is acceptable, but 455 or 101 or 222 is not acceptable. I think you can solve the problem in the following manner. There are 10 possible digits (0-9) that can be used in the first number. For the second number there are only 9 possible since you cannot repeat one of the first. For the third number there are only 8 possible since again you cannot repeat any of the first two. You multiply the possibilities for each digit to get our answer. 10 * 9 * 8 = 720 There are 720 possible combinations.

Assuming that the digits 0-9 are available, the answer is 5040. This is because for the first digit you can pick any of the original 10 digits. For the second you can have any otherdigit, so 9 possible choices. The third follows the same rule, so has 8 possibilities, and the forth digit, 7.This means that the total number of combinations is 10 x 9 x 8 x 7 = 5040

If the order of the numbers are important, then this is a simple combination problem. There are 10 possible numbers to choose from for the first number. Then there are 9 options for the second number. Then there are 8 options for the third, and so on. Thus, the number of possible combinations can be calculated as 10x9x8x7x6x5. This comes out at 151,200 possible combinations.

The number of 3-digit numbers with no repeated digits is simply 10x9x8 = 720, if you allow, for example, 012 as a 3-digit number. There are 10 digits, any of which might be the first digit. The second digit can be any digit except the digit that was used for the first digit, leaving 9 possibilities. The third digit then has 8 possibilities, since it can't be the same as the first or second digit. The actual number of possible area codes will be lower, because there are additional restrictions on the number combinations for a valid area code. For example, in North America (USA, Canada, etc.), the first digit of an area code cannot be 0 or 1 and the middle digit cannot be 9.

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.

For this type of "the largest number which..." questions, you need to advance from left to right, using the largest possible digit in each case. For the first two digits, that would be 9, for the third digit (the right-most digit), the largest digit which will make this possible is an 8.

To generate all possible three-digit combinations from the given pattern, you can start from 000 and increment by 1 until you reach 999. Here's an example: 000 001 002 ... 345 ... 998 999 Continue this pattern, incrementing each digit by 1 until all possible combinations have been generated.

There are 10 choices for the first number, 9 for the second and 8 for the third. 10*9*8=720 possible combinations.

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