How many unique ways are there to arrange the letters in the word HATTER? 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. 
 A: 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.
A: 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.
A: 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
$$
\frac{n!}{a_1! \cdot a_2! \cdot \ldots \cdot a_k!}
$$
A: 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 count of each letter factorial. In this case $$\frac{6!}{2!\cdot 1!\cdot1!\cdot1!\cdot1!}$$
