# How do you calculate permutations of a word?

How do you calculate permutations of a word?
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Step 1
For the first part of this answer, I will assume that the word has no duplicate letters.
To calculate the amount of permutations of a word, this is as simple as evaluating n!, where n is the amount of letters. A 6-letter word has $6!=6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1=720$ different permutations.
To write out all the permutations is usually either very difficult, or a very long task. As you can tell, 720 different "words" will take a long time to write out. There are computer algorithms and programs to help you with this, and this is probably the best solution.
Step 2
The second part of this answer deals with words that have repeated letters. One formula is $\frac{n!}{{m}_{A}!{m}_{B}!...{m}_{Z}!}$ where n is the amount of letters in the word, and ${m}_{A},{m}_{B},...,{m}_{Z}$ are the occurrences of repeated letters in the word. Each m equals the amount of times the letter appears in the word. For example, in the word "peace", ${m}_{A}={m}_{C}={m}_{P}=1$ and ${m}_{E}=2$. So the amount of permutations of the word "peace" is: $\frac{5!}{1!\cdot 1!\cdot 1!\cdot 2!}=\frac{5\cdot 4\cdot 3\cdot 2\cdot 1}{1\cdot 1\cdot 1\cdot 2\cdot 1}=60$
Step 3
I will go through two more examples, but I will ignore every instance of 1! since $1!=1$.
For the word "committee": ${m}_{C}={m}_{O}={m}_{I}=1$
${m}_{M}={m}_{T}={m}_{E}=2$
Permutations: $\frac{9!}{2!2!2!}=\frac{9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{\left(2\cdot 1\right)\cdot \left(2\cdot 1\right)\cdot \left(2\cdot 1\right)}=45,360$
For the word "cheese": ${m}_{C}={m}_{H}={m}_{S}=1$
${m}_{E}=3$
Permutations: $\frac{6!}{3!}=\frac{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{3\cdot 2\cdot 1}=120$