How many 7-digit numbers are there where all the digits are different and in increasing order? for example, the numbers 1345689 and 2456789 would be counted. (the first digit cannot be zero.) ?

For this problem, we use the Fundamental Counting Principle. You know that there are 7 digits in a number. In this principle, you have to multiply the possible numbers for every digit. If the first number cannot be zero, then there are 9 possible numbers. So, the value for the first digit is 9. The second digit could be any number but less of 1 because it was used in the 1st digit. So, that would be 10 - 1 = 9. The third digit must be the value in the second digit less than 1. That would be 9 - 1 = 8. And so on and so forth. The solution would be:

9*9*8*7*6*5*4 = 544,320 7-digit numbers