Permutations with repetition calculator

How many different “words” can you make by rearranging the letters of MATEMATIKA (the Czech word for maths)? If all ten letters were different, there would be 10! permutations. But the letter A appears three times in the word, M twice and T also twice. Swapping two As gives the same word, so such swapped duplicates must not be counted more than once.

Enter the total number of letters (items) as n, then how many times each letter is repeated. You do not need to fill in letters that appear only once. Leave the unused boxes at one, because 1! = 1 and so it does not change the result.

How it works

So n! is divided by the factorials of the repetition counts:

$$P^{\prime }(k_1,k_2,\dots )=\frac{n!}{k_1!\cdot k_2!\cdot \dots }$$

MATEMATIKA has ten letters. A appears three times, M and T twice each, and E, I and K once. If all the letters were different, there would be 10! = 3,628,800 orderings. But the three As can be swapped among themselves in 3! = 6 ways without changing the word, the two Ms in two ways, and the two Ts also in two ways. Divide:

$$P^{\prime }(3.2.2)=\frac{10!}{3!\cdot 2!\cdot 2!}=\frac{\mathrm{3,628,800}}{6\cdot 2\cdot 2}=\frac{\mathrm{3,628,800}}{24}=\mathrm{151,200}$$

So you can make 151,200 different words from the letters of MATEMATIKA. Most of them are nonsense, but MATEMATIKA, KAMATEMATI and TIKAMATEMA are among them.