The Math Forum

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


Date: 07/12/99 at 01:42:03
From: Katia Adge
Subject: Reducing the number of permutations using the probability of 

How can I reduce the number of arrangements of the word ARRANGEMENT by 
using the probability of an occurrence?

In this example, knowing that this word has 11 letters, 2 R, 2 A, 2 E, 
and 2 N, the number of permutations of the word ARRANGEMENT is 

Knowing that 2 N cannot be found next to each other, and the same for 
2 E and 2 A, do you know how can I use this information to reduce the 
number of arrangements of this word?

Thank you for answering my question.

Date: 07/12/99 at 07:15:47
From: Doctor Anthony
Subject: Re: Reducing the number of permutations using the probability 
of occurrences

If you want the number of arrangements when the 2 N's, 2 E's and 2 A's 
are not together, you first find the number of arrangements with one 
or more together and subtract this result from 11!/(2!2!2!2!).

To find the number of arrangements with 2 N's together, we join them 
together as one object, giving a total now of 10 letters.  

Number of arrangements with N's together = 10!/(2!2!2!).

Similarly with two E's = 10!/(2!2!2!) and 2 A's = 10!/(2!2!2!).

Now consider two N's together and 2 E's together. We now have 9 
objects and these can be arranged in  9!/(2!2!) ways. Similarly with 
E's and A's together and A's and N's together.

Finally with all three letters joined as pairs the number of 
arrangements is 8!/2!

Now using the notation n(N), n(E), n(E and A) to mean the number of 
arrangements with these letters joined together in pairs, we get from 
the inclusion-exclusion principle the formula

 n(N or E or A) = n(N) +n(E) +n(A) -n(N and E) -n(E and A) -n(A and N) 
                  + n(N and E and A)

                = 3 x 10!/(2!2!2!) - 3 x 9!/(2!2!) + 8!/2!

                = 1108800

And 11!/(2!2!2!2!) = 2494800.

So the number without the N's, E's or A's together is 
2494800 - 1108800 =  1386000  ways.

- Doctor Anthony, The Math Forum   
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- The Math Forum at NCTM. All rights reserved.