hi wazup hi wazup
450
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.
90
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.
9,000,000,000. And that is anywhere, not just in North America!
450
5
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.
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.)
a lot * * * * * Possibly, though that answer is relative. The correct answer is 900.
Without repeats there are 4 × 3 = 12 possible 2 digit numbers. With repeats there are 4 × 4 = 16 possible 2 digit numbers.
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
what is the least possible sum of two 4-digit numbers?what is the least possible sum of two 4-digit numbers?
56 combinations. :)
If you include 0000, ten thousand unique four digit codes are possible.
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.