Counting Valid Arrangements of aabbccde: A Comprehensive Guide

Introduction

One common problem in combinatorial mathematics is determining the number of valid arrangements for a set of letters with certain restrictions. For instance, the word aabbccde presents an interesting challenge: how many ways can we arrange these letters such that no two identical letters are adjacent? In this article, we explore this problem in detail, employing a systematic approach to arrive at the correct answer.

Total Arrangements

The first step in any combinatorial problem is to determine the total number of arrangements without any restrictions. Given the word aabbccde, we have a total of 8 letters, with a, b, and c each appearing twice, and d and e appearing once. The formula for permutations of a multiset is used in this scenario.

Total arrangements: (frac{8!}{2! times 2! times 2! times 1! times 1!})

Calculating this:

(8! 40320)

(2! 2) for each of a, b, and c
(text{Total arrangements} frac{40320}{2 times 2 times 2} 5040)

Arrangements with Alike Letters Together

The next step involves subtracting the cases where at least one pair of alike letters is together. This is done by treating each pair of alike letters as a single block.

Treating aa as a Block

The letters are now aa b b c c d e. This results in 7 items to arrange, with b and c appearing 2 times each:

Arrangements: (frac{7!}{2! times 2!} 1260)

Treating bb as a Block

The letters change to a a bb c c d e. Similar calculations yield:

Arrangements: (frac{7!}{2! times 2!} 1260)

Treating cc as a Block

The letters are a a b b cc d e. Again, we apply the formula:

Arrangements: (frac{7!}{2! times 2!} 1260)

Overlapping Cases

Some cases were subtracted twice, which necessitates adding them back. These involve scenarios where two pairs of letters are together.

Treating aa and bb as Blocks

The letters are aa bb c c d e. The number of arrangements is:

Arrangements: (frac{6!}{2!} 360)

Treating aa and cc

Here the letters are aa b b cc d e. Again, we calculate:

Arrangements: (frac{6!}{2!} 360)

Treating bb and cc as Blocks

The letters change to a a bb cc d e. The number of arrangements is:

Arrangements: (frac{6!}{2!} 360)

Treating aa, bb, and cc as Blocks

The letters are aa bb cc d e. The number of arrangements is:

Arrangements: (5! 120)

Applying Inclusion-Exclusion Principle

Finally, we apply the inclusion-exclusion principle to get the total number of valid arrangements:

Total with restrictions: (5040 - 1260 - 1260 - 1260 360 360 360 - 120)

Calculating this:

(5040 - 3780 - 1080 120)

( 5040 - 3780 - 1080 120 2220)

Conclusion: The number of ways to arrange the letters in aabbccde such that no two identical letters are adjacent is 2220.