# 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).
 Sci Advisor HW Helper Thanks P: 26,148 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.
Homework