My answer: $$\binom{5}{5}_{R}-75=126-75=51$$
However, I want to try this using inclusion/exclusion
Let U be set $\{1,2,3,4,5\}$ where we want to find the number of 5-digit combination allowing repetition. Find this value using inclusion/exclusion so that: |U|-|$A_{1}\cup A_{2}\cup A_{3}\cup A_{4}$|=|U|-75=51.
What I am completely stuck on is coming up with the correct |U| and |$A_{1}\cup A_{2}\cup A_{3}\cup A_{4}$|. This is what I need help on.
Please help me understand what I need to do. Thank you.
Suppose you want to find the opposite: the number of combinations that have at least one digit that occurs at least three times. Let $p_i$ be the set of 5-digit combinations that satisfy the property that the digit $i$ is repeated at least thrice. You can see that to find the opposite, you need to find $$|p_1 \cup p_2 \cup p_3 \cup p_4 \cup p_5|.$$ You can use the inclusion-exclusion principal on this. It seems intimidating, but notice that it's impossible for two or more distinct properties to occur at the same time (do you see why?) This will cut down on the computations significantly.
Once you find that, the answer you seek will be $$|S| -| p_1 \cup p_2 \cup p_3 \cup p_4 \cup p_5|,$$
where $S$ is the total amount of 5 digit combinations with no restriction.