Combinatorics, permutations of letters

In summary, the conversation discusses how to calculate the number of words with 5 letters that can be created from the letters in the word ALGEBRA, with each letter being used only once. The correct answer is 11 * 5!, which can be obtained by counting the number of words with no As, with one A, and with 2As separately. Another approach is to separate the cases of "at most one A" and "2 As".
  • #1
Hannisch
116
0

Homework Statement


How many "words" with 5 letters can be created from the letters in the word ALGEBRA? Each letter can be used only once.


Homework Equations





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!}{(7-5)!}[/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).
 
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  • #2
Hi Hannisch! :smile:

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. :wink:
 
  • #3
tiny-tim said:
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. :wink:
Slightly quicker still, just separate the cases "at most one A" (just as with all the other letters) and "2 As"
 
  • #4
oh yes! :biggrin:
 

1. What is combinatorics?

Combinatorics is a branch of mathematics that deals with counting and arranging objects or events in a systematic manner.

2. What are permutations?

Permutations refer to the different ways in which a set of objects or events can be arranged or ordered.

3. How do you calculate the number of permutations for a set of objects?

The number of permutations can be calculated by using the formula n! (n factorial), where n represents the number of objects in the set.

4. How is combinatorics used in real-life applications?

Combinatorics has various applications in fields such as computer science, genetics, and statistics. It is used to solve problems related to combinations, permutations, and probability.

5. What is the difference between a combination and a permutation?

A combination refers to the selection of objects without considering the order, while a permutation takes into account the order of the objects.

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