How do you calculate permutations of a word?

Dec 13, 2014

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*5*4*3*2*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.

The second part of this answer deals with words that have repeated letters. One formula is
(n!)/(m_A!m_B!...m_Z!)
where $n$ is the amount of letters in the word, and ${m}_{A} , {m}_{B} , \ldots , {m}_{Z}$ 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:
(5!)/(1!*1!*1!*2!) = (5*4*3*2*1)/(1*1*1*2*1) = 60

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: (9!)/(2!2!2!) = (9*8*7*6*5*4*3*2*1)/((2*1)*(2*1)*(2*1)) = 45,360

For the word "cheese":
${m}_{C} = {m}_{H} = {m}_{S} = 1$
${m}_{E} = 3$
Permutations: (6!)/(3!) = (6*5*4*3*2*1)/(3*2*1) = 120