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Q: How many different 7 digit phone numbers are possible if the first digit cannot be 0 or 1?

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450

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.

5

There are 900 possible three-digit numbers not beginning with 0. (Note, however, that this question does not accurately describe the restrictions on numbers that can be used as area codes.)

There are 30,240 different 5-digit numbers. Math: 10*9*8*7*6 1st digit has 10 possible choices (0-9) 2nd digit has 9 possible choices (one of the digits was used in the 1st digit) 3rd digit has 8 possible choices 4th digit has 7 possible choices 5th digit has 6 possible choices

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

If you include 0000, ten thousand unique four digit codes are possible.

what is the least possible sum of two 4-digit numbers?what is the least possible sum of two 4-digit numbers?

56 combinations. :)

Forming three three digit numbers that use the numbers 1-9 without repeating, the highest product possible is 611,721,516. This is formed from the numbers 941, 852, and 763.

9 x 10 x 10 x 5 = 45000

9*106 or 9 million.

450... There are 500 odd numbers from 000 to 999 inclusive. From 000 to 099, there are 50 odd numbers. 500 - 50 = 450.

There are 9 possible numbers for the first digit (one of {1, 2, ..., 9}); with 9 possible digits for the second digit (one of {0, 1, 2, ..., 9} which is not the first digit)); with 8 possible digits for the third digit (one of {0, 1, 2, ..., 9} less the 2 digits already chosen); This there are 9 × 9 × 8 = 648 such numbers.

6 possible 3 digit combonations

There are seven possible digits for the first digit and 6 digits for the second (minus one digit for the digit used as the first digit) and 5 options for the last digit (minus one again for the second digit) and then you just multiply them all together to get a total possible combination of 210 numbers that are possible.

If a digit can be repeated there are 5 x 5 x 5 = 125 possible numbers If a digit cannot be repeated there are 5 x 4 x 3 = 60 possible numbers.

There are 10 to the 10th power possibilities of ISBN numbers if d represents a digit from 0 to 9 and repetition of digits are allowed. That means there are 10,000,000,000 ISBN numbers possible.

a lot * * * * * Possibly, though that answer is relative. The correct answer is 900.

There are 4,500 combinations.

Add the two greatest possible four digit numbers. 9999 + 9999

Without restrictions, it was would numbers 000-000-0000 through 999-999-9999. So that would be 9,999,999,999 + 1 = 10 billion different 10-digit phone numbers. Ex: If there existed single digit phone numbers, there would be 10, because the digits are 0 through 9. If there existed only double digit phone numbers, then it would be 00 through 99 which would be 100 total two-digit numbers. Therefore the total possible combinations for an X digit phone number would be: 10^X

That makes:* 8 options for the first digit * 8 options for second digit * 10 options for the third digit * ... etc. Just multiply all the numbers together.

There are 720 of them. The three digit counting numbers are 100-999. All multiples of 5 have their last digit as 0 or 5. There are 9 possible numbers {1-9} for the first digit, There are 10 possible numbers {0-9} for each of the first digits, There are 8 possible numbers {1-4, 6-9} for each of the first two digits, Making 9 x 10 x 8 = 720 possible 3 digit counting numbers not multiples of 5.

The first digit can have 5 possible numbers, the second digit can have 4, the third 3, the fourth 2. 5

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