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

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

A. p = 2(l + w)p/2 - l = ww = p/2 - l b. while isolating the variable w in the equation, i first divided both sides by 2 and then i subtracted l from both sides. i hope this .