How many unique ways are there to arrange the letters in the word HATTER?

I can't wrap my head around the math to find the answer. I know that if they were all different letters the answer would be 6!. However, I know that these T's are going to overlap, so it won't be that.

I am trying to give myself examples like AAA, it can only be written once but if it was 3 different letters it would be 6 times instead. Somehow I need to get a 6/6, so that it can become 1.

If I try it with AAC, half of the permutations disappear. So it must be divided by 2 I guess. 6/2.

  • ABC AAC 1
  • ACB ACA 2
  • BCA ACA 2
  • BAC AAC 1
  • CAB CAA 3
  • CBA CAA 3

I kind of see a pattern here. Possible combinations if all letters were different factorial / Divide by the number of equal letters factorial, but still I am confused.

Explanation is appreciated.

The answer is 360.

  • $\begingroup$ If you're not limited to a single horizontal line, the answer is infinite. A vertical arrangement, in a circle, five letters scrunched together with the sixth stored in a cave in Botswana. The possibilities are endless, though I should be careful about using the words "infinite" and "endless" on a site frequented by mathematical pedants :-) $\endgroup$ – user1324 Jul 11 '16 at 8:39

Imagine one of the Ts is red and the other is blue. Then write out all 6!=720 arrangements. You will see that while they are all unique, you can create pairs where the only difference is the position of the red and blue Ts. Since they are identical in the original question, you must divide by the number of ways the Ts can be arranged. In this case, $2!=2$. So your answer is $$\frac{6!}{2!}$$

Generally you can write the answer as the total number of letters factorial divided by the number of each letter factorial. In this case $$\frac{6!}{2!\cdot 1!\cdot1!\cdot1!\cdot1!}$$

  • 3
    $\begingroup$ The equation on the bottom was helpful. Thanks. $\endgroup$ – David Lund Jul 10 '16 at 16:57

You have $6$ letters in total so that's $6! = 720$ total ways to permute the letters. In addition, the letter $T$ is repeated twice, so you have to divide out the number of ways to permute the $T$s, since a permutation of two identical letters doesn't matter. So you have

$$\frac{6!}{2!} = \frac{720}{2} = 360.$$

Hope this helps.


6 letters could be arranged in 6! ways. Since there are two same letters (T) each arrangement is counted twice - for example $HAT_{1}T_{2}ER$ and $HAT_{2}T_{1}ER$ are counted as 2 arrangements while it's the same. So number of ways should be half of 6! and the solution is $\frac{6!}{2}$.

There is pattern here, you're right. This situation is called permutations with repetition - if we have n objects where the first element is repeated $a_1$ times, the second $a_2$ times, ..., kth $a_k$ times ($a_1+a_2+...+a_k=n)$ then number of arrangements is $$ \frac{n!}{a_1! \cdot a_2! \cdot \ldots \cdot a_k!} $$

  • $\begingroup$ I have trust issues with the fact that if there are 720 combinations, half of them will go away in hatter. I can accept it, and you pointed out one (thanks), but I really want to see all the others so that I can believe. I want to do hundreds of them and just write them all out so I can really believe. On the other pattern proves it I guess, but it still feels unreal. $\endgroup$ – David Lund Jul 10 '16 at 17:00
  • $\begingroup$ If you have 3 same letters (A, for example) then there will be 6 (3!) arrangement: $A_1A_2A_3$, $A_1A_3A_2$, ... which should be count as 1: AAA. So number of arrangement should be divided by 6 (3!) $\endgroup$ – Nesh Jul 10 '16 at 17:05
  • $\begingroup$ This is a permutation of a multiset rather than a permutation with repetition since we rearranging the letters of a word of fixed length rather than forming words from an alphabet. $\endgroup$ – N. F. Taussig Jul 10 '16 at 17:33

You have 6 choices for where to place the H, 5 choices for the A, 4 choices for the E, and 3 choices for the R

(and then the 2 T's go in the remaining places).

This gives $\displaystyle 6\cdot5\cdot4\cdot3=360$ choices.


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