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560135

Micron: wished you had posted the logic. As I see it, the first can be one of 8 digits. The second number can be 0 or 1 but not the same as the first (9 to choose from). The third cannot be the same as either of the first two (8 to choose from), etc. Thus it would be 8*9*8*7*6*5*4=483,840

Q: How many 7 digit phone numbers are available if a digit can only be used once and the first number can not be 0 or 1?

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3 digit numbers are lessthan 4 digit numbers

1234567890

There are 10 possible numbers that can be chosen. Since the first digit can't be 0, there are 9 selections for the first digit. The second number can include a 0, but can't include the first number chosen, so there are 9 selections for the second digit. Then there are 8 digits for the next number, 7 digits for the next number, and so on. Total telephone numbers are: 9*9*8*7*6*5*4 = 544320 phone numbers.

There are 54 such numbers.There are 54 such numbers.There are 54 such numbers.There are 54 such numbers.

120

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3 digit numbers are lessthan 4 digit numbers

The first 8 digit number is 10,000,000, the last is 99,999,999 which means there are 90,000,000 8 digit numbers

Assuming that the first digit of the 4 digit number cannot be 0, then there are 9 possible digits for the first of the four. Also assuming that each digit does not need to be unique, then the next three digits of the four can have 10 possible for each. That results in 9x10x10x10 = 9000 possible 4 digit numbers. If, however, you can not use the same number twice in completing the 4 digit number, and the first digit cannot be 0, then the result is 9x9x8x7 = 4536 possible 4 digit numbers. If the 4 digit number can start with 0, then there are 10,000 possible 4 digit numbers. If the 4 digit number can start with 0, and you cannot use any number twice, then the result is 10x9x8x7 = 5040 possilbe 4 digit numbers.

Zero.

1234567890

The smallest three-digit number divisible by the first three prime numbers (2, 3, and 5) and the first three composite numbers (4, 6, and 8) is 120.

The answer will depend on how many digits there are in each of the 30 numbers. If the 30 numbers are all 6-digit numbers then the answer is NONE! If the 30 numbers are the first 30 counting numbers then there are 126 combinations of five 1-digit numbers, 1764 combinations of three 1-digit numbers and one 2-digit number, and 1710 combinations of one 1-digit number and two 2-digit numbers. That makes a total of 3600 5-digit combinations.

Let's look at this digit-by-digit: The first digit can be any number 1-9: 9 choices The second digit can be any number 1-9 except the one that the first digit is: 8 choices The third digit can be any number 1-9 except the ones chosen by the first and second digits: 7 choices 9*8*7 = 504 total numbers

There are 10 possible numbers that can be chosen. Since the first digit can't be 0, there are 9 selections for the first digit. The second number can include a 0, but can't include the first number chosen, so there are 9 selections for the second digit. Then there are 8 digits for the next number, 7 digits for the next number, and so on. Total telephone numbers are: 9*9*8*7*6*5*4 = 544320 phone numbers.

There are 54 such numbers.There are 54 such numbers.There are 54 such numbers.There are 54 such numbers.

120

10 because 10 is the first 2 digit number out of all numbers.