Arrangements With Repeated Letters

Why Repeated Letters Change the Count

If every letter in a string is different, arranging its \(n\) letters gives \(n!\) possible orders. When some letters repeat, however, exchanging two copies of the same letter does not produce a visibly different string. Counting all \(n!\) orders as distinct would therefore overcount. The key is to identify the repeated letters and divide out the number of exchanges that leave the string unchanged.

For this tutorial, an arrangement means a string formed by using every letter exactly as many times as it appears in the starting collection. We count different visible strings, not different ways of labeling identical copies. The resulting strings need not be dictionary words. Unless stated otherwise, case is ignored, and spaces or punctuation are not included as letters.

A letter appearing only once has multiplicity 1, and \(1!=1\), so it does not change the denominator. It is still useful to keep track of such letters when checking that the multiplicities add to the total number of letters. Factorials count ordered arrangements of distinct objects; the denominator corrects that count for exchanges among identical copies.

Why the Formula Works

Temporarily imagine that identical letters have been given labels. For instance, two copies of A could be called \(A_1\) and \(A_2\). With all labels in place, there are \(n!\) orders. Removing the labels turns these labeled orders into visible strings, and many labeled orders become the same visible string.

For any one visible string, its \(r_1\) copies of the first letter can be assigned their labels in \(r_1!\) ways. Independently, its \(r_2\) copies of the second letter can be labeled in \(r_2!\) ways, and likewise for every other type. Thus every visible string comes from exactly \(r_1!r_2!\cdots r_k!\) labeled orders. Dividing the total labeled orders by this common number gives the formula:

The division is exact because the labeled orders are partitioned into groups: each group consists of all labelings that produce one particular visible string, and every such group has the same size. This also explains why dividing by \(r_i!\) is necessary for each repeated letter, not just for one of the repeated letters.

Several Letters Can Repeat

The same procedure applies when several different letters repeat. Make a frequency list first, check that its counts sum to the total length, and include one factorial in the denominator for each letter type. It is not enough to notice only the most frequent letter: every repeated group creates its own overcount.

A Systematic Method and a Small-String Check

For a new word or character string, use a fixed sequence of checks. This helps prevent both incorrect totals and missing denominator factors.

A short string makes the distinction between labeled and visible arrangements easy to inspect. Take AABBC, with 5 letters: A occurs twice, B occurs twice, and C occurs once. The formula gives:

A direct position check confirms this count. Choose 2 of the 5 positions for A, then 2 of the remaining 3 positions for B. The remaining position is C, giving \(\binom{5}{2}\binom{3}{2}=10\cdot3=30\). The two approaches agree because each arrangement is determined completely by the positions occupied by each letter type.

Common Mistakes and Why the Count Matters

Common mistake: A student sees four copies of one letter and reasons that the number of arrangements is \(n!/4!\), even though another letter is also repeated. That division corrects only for exchanging the first set of identical copies. Exchanges within every other repeated group still create overcounting.

Correction: List all letter multiplicities before calculating. If a word has 10 letters with multiplicities 3, 3, 2, 1, and 1, use \(10!/(3!3!2!1!1!)\), not \(10!/3!\). The multiplicities must sum to 10, and each repeated letter type supplies a denominator factor.

Another error is treating identical copies as distinguishable because they occupy different positions. Positions matter in describing a final string, but exchanging two copies of the same letter leaves the displayed string unchanged. For example, labeling the As in AAB as \(A_1\) and \(A_2\) creates two labeled versions of each visible arrangement, not two different visible arrangements. That is exactly the overcount the denominator removes.

These counts are useful whenever possible outcomes are strings formed from a fixed collection of symbols. The formula gives a precise total without listing each arrangement, and that total can then serve as a denominator in a probability calculation when the relevant outcomes are equally likely. The count alone does not establish that outcomes are equally likely; that must come from the way the random process is defined.