Discrete Math, anagram combinatorics. Find the number of anagrams for the word "ALIVE" so that the letter "A" is before the letter "E" or the letter "E" is before the letter "I". By before we mean any letter previous, not just immediately before.

Discrete Math, anagram combinatorics
Find the number of anagrams for the word "ALIVE" so that the letter "A" is before the letter "E" or the letter "E" is before the letter "I". By before we mean any letter previous, not just immediately before.
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Dominique Ferrell
Step 1
You have the word AEI and should put somewhere between its letters 2 another letters.
Step 2
There are 4 places for the first letter, _A_E_I_, and 5 places for the second letter after inserting the first. So the answer is $4\cdot 5=20$
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Emmanuel Pace
Step 1
There are 5! ways to arrange the given letters, if there are no restrictions. But there are restrictions.
We want A before E or E before I (or both). Call such an arrangement good.
We first count the bad arrangements, in which A is after E and E is after I, so the letters come in the order I, E, A, with possibly stuff between these letters.
Step 2
The 3 locations of our key letters can be chosen in $\left(\genfrac{}{}{0}{}{5}{3}\right)$ ways. Once we have chosen therse 3 spots, which of these spots is occupied by our letters is determined. And now we can arrange the remaining 2 letters in the 2 empty slots in 2! ways, for a total of $\left(\genfrac{}{}{0}{}{5}{3}\right)\left(2!\right)=20$ bad arrangements.
It follows that the number of good arrangements is $5!-20$, that is, 100.