8*10*10=800
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.
27 three digit numbers from the digits 3, 5, 7 including repetitions.
ten factorial = 10! = 3,628,800
from 100 to 999 it is 900
24 of them. So we have a 2 or 6 in the unit position, therefore (2)(4)(3) = 24 even three-digit numbers, can be formed.
81
There are 9 x 10 x 10 x 10, or 9000 four-digit number that can be formed if the leading digit cannot be zero.
9,000,000 if there are no other requirements.
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.)
1 set
It is 120 if the digits cannot be repeated.
Using your rule - essentially, it's the simplified sum of 9999999-2000000 which is a total of 7999999 numbers !
7
There are 7*94 = 45927 such numbers.
5040 different 4 digit numbers can be formed with the digits 123456789. This is assuming that no digits are repeated with each combination.
9999
Any 3 from 10 in any order = 10 x 9 x 8 = 720It depends on what other restrictions you place on the numbers. For example, in North America (country code +1 = USA, Canada, etc.), area codes cannot have 0 or 1 as the first digit, and cannot have 9 as the second digit.