# Combinatorics, permutations of letters

by Hannisch
Tags: combinatorics, letters, permutations
 P: 116 1. The problem statement, all variables and given/known data How many "words" with 5 letters can be created from the letters in the word ALGEBRA? Each letter can be used only once. 2. Relevant equations 3. The attempt at a solution I know the answer to this (1320) and I know how I got to it (which I'll describe in a minute), but I know that while my way of getting it is right, it's. not the easiest, and I can't seem to figure out how on earth they want me to do it. I know that choosing 5 things from a set with 7 elements can be done in $\frac{7!}{(7-5)!}$ ways. I also know that if I were to use all the letters of ALGEBRA the answer would be $\frac{7!}{2!}$ since one letter is used twice. In this case these happen to be exactly same, but I can't seem to put these principles together to get a correct answer. Help, please? The way I solved it was more of a brute force solution. The correct answer is 11 * 5!, and I got to that conclusion by reasoning that 5! of the words are made by the letters LGEBR, which have no recurring letters. Then I "exchanged" one of the letters from that to an A, one at a time, and then to both As. This can be done in 11 ways: LGEBR AGEBR LAEBR LGABR LGEAR LGEBA AAEBR LAABR LGAAR LGEAA AGEBA And all of them can be chosen in 5! different combinations. I know this is the correct answer (it's an online based homework, I've inputted this and it says I'm right).
 PF Patron HW Helper Sci Advisor Thanks P: 25,474 Hi Hannisch! That certainly works. But quicker would be to split the problem into three … count separately the number of words with no As, with one A, and with 2As.
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P: 7,843
 Quote by tiny-tim But quicker would be to split the problem into three …count separately the number of words with no As, with one A, and with 2As.
Slightly quicker still, just separate the cases "at most one A" (just as with all the other letters) and "2 As"

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