well you would probably have to mutiply 9 times ten times ten times ten, because you have four spaces for any number up to nine and zero can't be used in the first digit. My answer would have to be 9000, although I'm only in eighth grade. Anyone who has time to write out every possibility, though, has no life.
There are 22680 numbers, excluding those with leading 0s.
There are 5460 five digit numbers with a digit sum of 22.
To find the even two-digit numbers where the sum of the digits is 5, we need to consider the possible combinations of digits. The digits that sum up to 5 are (1,4) and (2,3). For the numbers to be even, the units digit must be 4, so the possible numbers are 14 and 34. Therefore, there are 2 even two-digit numbers where the sum of the digits is 5.
2
There are 5760 such numbers.
9876543120
9999999999 100000000
Any pair of digits (not including 0), can be used to generate 14 four-digit numbers. If one of the digits is 0, only seven will start with a non-zero digit.
1
Exactly 2: 243 of them.
There are 22680 numbers, excluding those with leading 0s.
that question cant be answered unless it was 3 odd-digits or 4 odd-digits with 1 even digit i think.
Zero; all six digit numbers have six digits. Dude.. i m asking 6 digit numbers, containing only 4 different digits..!! e.g.:123412.. with only 2 digits repeated..!! My guess is: 10*9*8*7*4*3 but I'm also guessing that is wrong- sorry!
With 123 digits you can make 123 one-digit numbers.
There are 5460 five digit numbers with a digit sum of 22.
5040 different 4 digit numbers can be formed with the digits 123456789. This is assuming that no digits are repeated with each combination.
There are 9 digits that can be the first digit (1-9); for each of these there is 1 digit that can be the second digit (6); for each of these there are 10 digits that can be the third digit (0-9); for each of these there are 10 digits that can be the fourth digit (0-9). → number of numbers is 9 × 1 × 10 × 10 = 900 such numbers.