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. 
Any help, totally stumped on this question.
 A: 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.
The $3$ locations of our key letters can be chosen in $\binom{5}{3}$ 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 $\binom{5}{3}(2!)=20$ bad arrangements.
It follows that the number of good arrangements is $5!-20$, that is, $100$.
A: You have the word $AEI$ and should put somewhere between its letters $2$ another letters. 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 $4\cdot 5 = 20$.
A: An approach that maybe scales better then just trying to enumerate the possibilities in a systematic way is the following. You probably know that there are $5!=120$ anagrams without restriction. In any such anagram the three vowels occur in some order among the $3!=6$ possible orders; the orders AEI, AIE, EAI, EIA and IAE satify you requirement, but IEA does not. Choosing an anagram and permuting only its vowels (leaving the other letters in place) gives a packet of $6$ anagrams, each one with $5$ solutions to you problem. The number of such good anagrams equals five times the number of packets, which is 
$$5\times\frac{120}6
=5\times\frac{5\times 4\times3\times2\times1}{3\times2\times1}
=5\times5\times 4=100.
$$
