Word Rearrangements Fundamentals

Rearranging the letters of a word is a permutation problem in disguise. When every letter is different, the count is a plain factorial, but repeated letters change the count because swapping two identical letters never produces a new word. This lesson builds the multinomial formula that handles any mix of repeated letters.

What you will learn

• Count arrangements of a word with all distinct letters
• Adjust the count for repeated letters
• Apply the multinomial arrangement formula

Keep paper and a pencil nearby. Try each problem before opening a hint, and mark sections reviewed as you work through them.

Arranging a word with all distinct letters

A word rearrangement asks how many different strings of letters can be formed by reordering the letters of a given word. When every letter in the word is different from every other letter, this is exactly the same problem as arranging $n$ distinct objects in a row, and the count is $n!$.

For example, the word CRISP has 5 letters, all different, so there are $5!=120$ ways to rearrange them. Each of the 120 outputs is a distinct string, even though most of them are not real English words, because the counting question only cares about the order of the letters, not whether the result is meaningful.

Why repeated letters change the count

When a letter repeats, treating the repeated copies as if they were distinguishable overcounts the arrangements. Imagine labeling the two L's in LEVEL as $L_1$ and $L_2$ and the two E's as $E_1$ and $E_2$: this labeled version has $5!=120$ arrangements. But swapping $L_1$ and $L_2$ with each other produces the exact same real word, and the same is true for swapping $E_1$ and $E_2$, so every real arrangement of LEVEL was counted multiple times in that labeled total.

Each group of identical letters contributes a factor equal to the number of ways to permute that group among itself. Dividing the labeled total by those factorials removes the overcounting and gives the true number of distinct words.

The multinomial arrangement formula

Combining these ideas gives the multinomial formula: if a word has $n$ letters total, and its distinct letters repeat $r_1, r_2, \ldots$ times, then the number of distinct arrangements is $\dfrac{n!}{r_1!\,r_2!\cdots}$. A word with no repeated letters is the special case where every $r_i=1$, and every factor in the denominator equals $1!=1$, recovering the plain $n!$ count from the first section.

Applying the formula only takes two steps: count how many letters the word has in total, then count how many times each distinct letter repeats. Once every repeated group's factorial is multiplied into the denominator, the division gives the exact count with no further adjustment needed.

See the reasoning unfold

Pause after each step and predict what comes next.

Common mistakes to catch

• Using $n!$ for a word with repeated letters without dividing by the repeated letters' factorials, which overcounts identical-looking rearrangements as if they were distinguishable.
• Miscounting how many times a letter repeats, especially in longer words, which changes the denominator of the multinomial formula.
• Forgetting to include every repeated letter group in the denominator when a word has more than one letter that repeats.

Try with support

Guided practice

Choose a strategy, take a first step, and use the hints when you need direction.

Put it into practice

Independent practice

Work on your own first. If you reveal a solution, work through it and mark it reviewed.

Bring it together

• A word with all distinct letters is arranged in $n!$ ways, just like any $n$ distinct objects.
• Each group of identical letters divides the count by that group's factorial, since swapping identical letters produces the same word.
• The multinomial formula $\dfrac{n!}{r_1!\,r_2!\cdots}$ divides by one factorial per repeated-letter group at once.