# Arranging the letters of the word ACCESSORIES given two constraints.

In how many different ways can we arrange the letters of the word ACCESSORIES so that no two similar letters are adjacent to each other, as well as no two vowels are adjacent to each other?

Examples of valid words: ACESORISECE, ESECOSIRECA, CEIOSARESEC.

Examples of invalid words: ACCESOSRISE (two similar letters are adjacent "CC"), CACESOSRIES (two vowels are adjacent "IE"), ACECOIRSSES (two similar letters are adjacent "SS" and two vowels are adjacent "OI").

I only know how to count the number of words that can be formed using the letters; A,C,C,E,E,I,O,R,S,S,S. The number of words $$=\frac{11!}{1!\times2!\times2!\times1!\times1!\times1!\times3!}=1663200$$ words, but I do not know how to exclude the invalid words.

The difficulty is: Excluding a word more than one time, for example excluding the word CACESOSRIES because it has two adjacent vowels "IE", then (by exchanging the positions of the two E's that leads to the same word actually) we again exclude that word because it has two adjacent vowels "IE", which is wrong. And I have other difficulties and confusions.

Any help would be appreciated. THANKS!

• Are you familiar with the Inclusion-Exclusion Principle? Sep 1 '19 at 10:06
• @N.F.Taussig No dear. Sep 1 '19 at 11:06

Since there are $$5$$ vowels over $$11$$ letters, we first arrange them so that we have no adjacent vowels.

We distinguish four classes of arrangements (here $$X$$ is a place for a vowel and $$Y$$ is a place for a consonant):

(i) $$1$$ with no adjacent consonants: $$YXYXYXYXYXY$$

(ii) $$4$$ with 3 adjacent consonants: $$X\color{blue}{YYY}XYXYXYX, XYX\color{blue}{YYY}XYXYX, XYXYX\color{blue}{YYY}XYX, XYXYXYX\color{blue}{YYY}X$$

(iii) $$10$$ with 2 adjacent consonants: $$\color{blue}{YY}XYXYXYXYX,YX\color{blue}{YY}XYXYXYX,YXYX\color{blue}{YY}XYXYX, YXYXYX\color{blue}{YY}XYX, YXYXYXYX\color{blue}{YY}X, X\color{blue}{YY}XYXYXYXY,XYX\color{blue}{YY}XYXYXY,XYXYX\color{blue}{YY}XYXY, XYXYXYX\color{blue}{YY}XY, XYXYXYXYX\color{blue}{YY}$$

(iv) $$6$$ with 2 couples of adjacent consonants: $$X\color{blue}{YY}X\color{blue}{YY}XYXYX, X\color{blue}{YY}XYX\color{blue}{YY}XYX, X\color{blue}{YY}XYXYX\color{blue}{YY}X,XYX\color{blue}{YY}X\color{blue}{YY}XYX,XYX\color{blue}{YY}XYX\color{blue}{YY}X,XYXYX\color{blue}{YY}X\color{blue}{YY}X$$

Hence, the total number of valid arrangements is $$\underbrace{\frac{5!}{2}}_{\text{vowels}}\cdot \left(\underbrace{1\cdot\left(\frac{6!}{2!3!}\right)}_{\text{case (i)}}+ \underbrace{4\cdot31}_{\text{case (ii)}}+ \underbrace{10\cdot\left(\frac{6!}{2!3!}-16\right)}_{\text{case (iii)}} +\underbrace{6\cdot\left(\frac{6!}{2!3!}-28\right)}_{\text{case (iv)}} \right)=48960.$$ Please fill the details of the above computation and let me know if you need further help.

• I have done this with a slightly different method (although effectively the same) and got a different answer. Am I not correct in thinking that case ii should be 4 times 3? Sep 1 '19 at 12:24
• @S.Dolan Do you mean $4\cdot 3$ instead of $4\cdot 31$? Sep 1 '19 at 12:40
• Yes ( I don't seem to be able to send 'yes' on its own!) Sep 1 '19 at 12:48
• @S.Dolan Note that $YYY$ can be $SCR$ permuted, then fill the remaining $Y$ with $SSC$ which is already $3!\cdot 3>3$. Sep 1 '19 at 12:50
• Yes sorry, my mistake - trying to compare my solution with yours, I was looking at the wrong term. Sep 1 '19 at 12:57

First obtain the number of ways of arranging $$CCRSSS$$.

a) No constraints $$60$$

b) If the first two letters are equal $$16$$

c) If the first two letters are equal and the next two letters are also equal $$4$$

d) If the first three letters are equal $$3$$

Now let V stand for a vowel and C for a consonant or several consecutive consonants.

There are four possible arrangements.

1. CVCVCVCVCVC

2. CVCVCVCVCV and its reverse, where one 'C' is a pair of consonants

3. VCVCVCVCV, where two 'C's are both pairs of consonants

4. VCVCVCVCV, where one 'C' consists of three consonants

The vowels

These can be arranged in 60 ways. We now only have to deal with just the consonants.

Case 1

This is a) and so there are 60 arrangements.

Case 2

For each of this and its reverse,there are five ways of choosing the place for the pair of consonants. We must now subtract the cases b) from those of a) to obtain 44 arrangements of the consonants. A total of 440.

Case 3

There are six ways of choosing the two places for the pairs of consonants. We now have 60-16-16+4=32 arrangements of the consonants. A total of 192.

N.B. Note that we have to add back in answer c) since it has been counted in both the 16s.

Case 4

There are four ways of choosing the places for the triple of consonants. The first two of these consonants must be unequal and the second and third of these must also be unequal. We now have 60-16-16+3=31 arrangements of the consonants. A total of 124

N.B. Note that we have to add back in answer d) since it has been counted in both the 16s.

The total!

This is 60+440+192+124 multiplied by 60 i.e. 48960.

• (+1) Now the counting is correct. Sep 1 '19 at 13:26