Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Possible Letter Arrangements


Date: 07/05/2000 at 04:42:11
From: David Stanley
Subject: Permutations

Is there a formula I can use to get the number of possible 
arrangements of five, six, or seven different letters? I can do it the 
long way by writing them out, but I know there must be another 
quicker, foolproof way. Four letters, ABCD, can be arranged in 24 
different patterns. What's the trick to doing it quick?


Date: 07/05/2000 at 13:14:25
From: Doctor Rick
Subject: Re: Permutations

Hi, David.

How many choices are there for the first letter? Four, of course: it 
can be A, B, C, or D.

Suppose you choose A. Then how many choices are there for the second 
letter? Only three this time; it can be B, C, or D.

Suppose you choose to put B in the second position. Then how many 
choices are there for the third letter? There are only two choices 
left: the third letter can be C (with the last letter being D), or 
vice versa (D, then C.)

Now, it doesn't matter which letter you choose to put in the second 
position; for each of the three choices (B, C, or D), you can make
two different arrangements:

     Choose B --> ABCD or ABDC
     Choose C --> ACBD or ACDB
     Choose D --> ADBC or ADCB

So the total number of arrangements you can make with A in the first 
position is 3 * 2 = 6.

Go back to the first letter. There are four choices. Again, it 
doesn't matter which choice we make, there will be 6 arrangements we 
can make in each case. The total number of arrangements we can 
make is 4 times 6 = 24.

In the time it took us to go through this, you could have written out 
all 24 words easily. But now you can see the rule that makes it easy 
to do similar problems - such as the number of arrangements for 5 
letters. Look at the problem as a series of choices that you make one 
at a time. If the number of choices you have at one step doesn't 
depend on the choices you made previously, then you can just multiply 
the number of choices you have at each step. In your problem, you have 
4 choices for the first letter, 3 choices for the second letter, and 
2 choices for the third letter, so the total number of choices is 
4 * 3 * 2 = 24.

Does this make sense? If you're not sure, go ahead and write out all 
the arrangements. Organize them the way I did above, and it might help 
you to see why you can multiply the numbers of choices.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Permutations and Combinations

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/