Could someone help me with this?
I think I need to use inclusion/exclusion... so something like:How many 6 character lists can you make, where A-Z can be used for the first 3 positions and the digits 0-9 can be used for the last 3 positions. Important: The letter O and the digit 0 cannot be used in the same list.
number of all possible permutations (26^3 * 10^3) minus number of bad lists (lists with both O and 0).
I'm not sure how to calculate the number of bad lists, tho.
Thanks for your help!
Thanks for your reply! We actually used that method in class, but I'd like to know how to do it this way as well.
I've made a little more progress, I think...
Let X be the set of all possible permutations.
That is, |X| = (26)^3 * (10)^3
Let A be the set of permutations containing at least one 0.
That is, 26^3 * 1*10^2
Let B be the set of permutations containing at least one O.
That is, 1*26^2 * 10^3
Would I now need to find the intersection of A and B and subtract that from X? How would I find that intersection?
Your counts of the permutations containing at least one 0 etc. are wrong. Remember that (1) the 0 may occur in any of 3 places (2) "at least one" includes 2 or 3 as well as 1.
If you really want to find the number of 3-digit numbers containing at least one 0, it's probably easiest to find the number of 3-digit numbers which do not contain any 0 and subtract that from 10^3.
The same approach applies to the problem of finding the number of 3-letter "words" containing at least one O.