Number of unique possible ways to arrange the letters U, T, T, E, R, A, L, L How many possible unique ways are there to arrange the letters: U, T, T, E, R, A, L, L, all eight of them taken at a time?
If these were all unique letters, you could simply do 8! to find the answer. However, two pairs (T, T and L, L) in these are the same, which means that 8! in this case would give you duplicate permutations.
The answer would be less than 8! But I am clueless as to how you actually approach the answer. Thus far, I've tried
6 (since there are 6 unique letters, let's say you choose T first) * 6 (since there are still 6 unique letters, another T) * 5 (now 5 unique letters, chose L) * 5 (L) * 4 * 3 * 2 * 1 to get 21600, but this isn't the right answer. 
 A: We have $1$ A, $1$ E, $2$ Ls, $1$ R, $2$ Ts, and $1$ U to arrange in eight positions.  
Choose two of those eight positions for the Ls.  Choose two of the remaining six positions for the Ts.  The remaining four positions can be filled with the remaining four letters in $4!$ ways.  Hence, the number of distinguishable permutations of the letters is 
$$\binom{8}{2}\binom{6}{2}4! = \frac{8!}{2!6!} \cdot \frac{6!}{2!4!} \cdot 4! = \frac{8!}{2!2!}$$
One factor of $2!$ in the denominator represents the number of ways we could permute the Ls among themselves within a given arrangement without producing an arrangement that is distinguishable from that arrangement.  The other factor of $2!$ in the denominator represents the number of ways we could permute the Ts among themselves within a given arrangement without producing an arrangement that is distinguishable from that arrangement.
A: If we have elements (here: letters) $\{a_1, a_2, ... ,a_n\}$ and each of them repeats $\{k_1, k_2, ...,k_n\}$ times, then the formula is:
$$\frac{(k_1+k_2+...+k_n)!}{k_1!k_2!...k_n!}$$
We have elements U, T, E, R, A, L, they repeat 1,2,1,1,1,2 times. So, $(k_1+k_2+...+k_n)!=(1+2+1+1+1+2)!=8!$
$$k_1!k_2!...k_n!=1!2!1!1!1!2!=2!2!$$Finally, we have $$\frac{8!}{2!2!}$$
