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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.

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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, $100$.

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I get the 5 choose 3 portion, but how do you know that every combination of the 5 choose 3 is a bad arrangement? It seems like most of what comes of the 3 elements is an acceptable arrangement. Maybe I'm just missing something? P.S. Thanks a lot for the help, all of you guys answering, very much appreciated. – Michael Y Apr 24 '13 at 6:53
Imagine we have $5$ consecutive slots where the letters are supposed to go. To make a bad word, we choose $3$ of these slots, and put the letters I, E, and A in the slots, in that order. Whatever we do with the other $2$ letters, this is a bad word. And all bad words are produced in this way. – André Nicolas Apr 24 '13 at 7:00
AH! Just realized the in that order part. Totally got it now. Thanks a lot. – Michael Y Apr 24 '13 at 7:07
You are welcome. The viewpoint I described is useful later, when one looks for example at the binomial distribution. – André Nicolas Apr 24 '13 at 7:10

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 $4\cdot 5 = 20$.

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You seem to have misread the question (as I did). The condition as A before E or E before I. – Marc van Leeuwen Apr 24 '13 at 5:34
Oops. Anyway, there are already 2 good answers. – Harold Apr 24 '13 at 5:55

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 $$5\times\frac{120}6 =5\times\frac{5\times 4\times3\times2\times1}{3\times2\times1} =5\times5\times 4=100. $$

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