Fundamental principles of counting

Question

Of all numbers between 10,000 and 99,999, inclusive how many.

A- Do not contain the digit 5?

B- Do contain the digit 5?

C- Are odd and contain no digit more than once?

D- Have no two consecutive digits the same?

I've worked part D, and I got 9^5 would be the answer, but I just want help on A-->C.

Answer 1

You're going to have to use the fundamental principle of counting. Since all of these numbers will be between $10,000$ and $99,999$, there are $9$ possible first digits, and $10$ possible digits for each other place value.

Problem A: If none of them contain the number $5$, then there are $8$ possibilities for the first digit and $9$ for the others, for a total of $8*9^4$ numbers.

Problem B: If all of them contain $5$ and there are a total of $90,000$ numbers between $10,000$ and $99,000$ inclusive, the number containing $5$ is $90,000$ minus the number that don't contain $5$, which is $90,000-8*9^4$.

Answer 2

Some hints:

For A, pick each digit such that it's not a $5$. The leftmost digit is handled differently than the other four.

For B, use the fact that the set of numbers in A and this set are mutually exclusive.

For C, pick an odd rightmost digit. Pick a tens digit that's different (may or may not be odd). Pick a hundreds digit that's different than the other two. And so forth.