None. The sum of one digit can't be twice the size of the digit.
Excluding numbers which begin with 0, there are 67,200 of them.
Only one . . . . . 18
24.
120 5-digit numbers can be made with the numbers 12345.
5*4*3*2 = 120
There are 360 of them.
9,000,000 if there are no other requirements.
That would be the numbers in the form "32x" (where "x" can be any digit). In other words, ten numbers.
As no digit is used twice then you have 4 choices for the first digit, but only 3 choices for the 2nd digit, 2 for the 3rd and just one for the 4th. 4 x 3 x 2 x 1 = 24 combinations.
How many two digit numbers are there in which the tens digit is greater than the oneβs digit ?
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.