8+9-1-2-3=11
1. The numbers 1 through 99 are 99 different numbers; each one occurs once.
998,000
Assuming each number can be used more than once.... 2401 possible combinations.
it cant
860 - 579 = 281 and 986 - 705 = 281
1. The numbers 1 through 99 are 99 different numbers; each one occurs once.
998,000
120, if you use each number once.
1346
24 = 7 * 8 / 2 - 4
There are 10,000 possible combinations, if each number can be used more than once.
Assuming each number can be used more than once.... 2401 possible combinations.
Choose from 4 for the first number, 3 for the second number, and 2 for the third number. Therefore there are 4*3*2 = 24 three digit numbers can be formed from 3769 if each digit is used only once.
dude its 8
720 assuming that each number can be used only once.
it cant
There should be 125 combinations, but if you can only use the numbers once for each 3 digit number, then it should be 120. It's 125 because if the digits could be used more than once for each number then its the number on possible numbers (so there's 5 in this case, 2, 3, 5, 8, and 9) to the power of how big the digit is (in this case it's 3) It's 120 because if the digits could only be used once in each number then the combinations would be the number of possible numbers (again, 5) factorial, so 5!