You bought six numbers at your local hardware store. The numbers are 0, 1, 2, 3, 4, 5. I got this question and can't crack it. Any help will be appreciated.
You bought six numbers at your local hardware store. The numbers are 0, 1, 2, 3, 4, 5.
a) How many 6 digit house numbers would be even?
(note: 0351 is not considered a house number and you only use each number once).
b) Using the same digits as above, how many 4 digit numbers smaller than 3500 could
you form if repetition was not allowed? (note: 0351 is not considered a house number).
 A: For (a), if you asked whether a random permutation was a valid house number, you have two "bad" things that could happen: it could start with zero, or it could end with an odd number.  You want neither of these to happen, so use the principle of inclusion/exclusion.
Let $N$ be the total number of permutations of your digits, $N_1$ and $N_2$ be the number of ways each of these two bad things could happen, and $N_{12}$ the number of ways both bad things happen.  Then the number of valid house numbers is
$$N-N_1-N_2+N_{12}.$$
In your case, $N=6!$, $N_1=5!$, $N_2=3(5!)$, and $N_{12}=3(4!)$, so the answer is $312$.
For (b), consider two cases according to the first digit.  If the first digit is less than $3$, then the remaining $3$ digits can be chosen however you like, so the count is $3\cdot 5\cdot 4\cdot 3$ in this case.  If the first digit equals $3$, then there are only $4$ choices for the next digit (as it can't be $5$), and then $4$ and $3$ choices respectively for the last two digits.  The answer is then $3\cdot5\cdot 4\cdot 3 + 4\cdot 4\cdot 3=288$.
A: In the context of your question, a 6-digit house number would be even if it ends with a 0, 2 or a 4. For each of these, there are 5! house numbers. So, you should have 3(5!) 6 digit house numbers in total.
For part (b), you will have to clarify the note about "0351 not considered a house number" and why it is relevant to how many 4 digit numbers there are.
