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Combinatorics, permutations of letters 
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#1
Dec1712, 12:22 PM

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 [itex]\frac{7!}{(75)!}[/itex] ways. I also know that if I were to use all the letters of ALGEBRA the answer would be [itex]\frac{7!}{2!}[/itex] 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). 


#2
Dec1712, 02:23 PM

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P: 26,160

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. 


#3
Dec1712, 11:01 PM

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