Proof of formula for permutation without repeating elements? My book wants me to find every unique word of "ABRAKADABRA" and since I know the formula:
Factorial of the set length divided by the product of the integer values of how many time an element has repeated itself, in this case = 11!/(5!*2!*2!)
Though I realise I can't rationalise this in my head. I tried painting a choice tree of something like the word "ABBA" but did not get a better understanding.
 A: There are $11$ positions. Choose the positions that will be $A$. There are $\dbinom{11}{5}$ ways to choose. Regardless of the positions chosen for the $A$'s, there are $6$ positions remaining. From among them, choose positions for $B$. There are $\dbinom{6}{2}$ ways to place them. Regardless of how the A's and B's are placed, there are 4 positions remaining. Choose how to place the single $D$: There are $\dbinom{4}{1}$ way to place the $D$. Then, there are $\dbinom{3}{1}$ way to place the $K$, and finally $1$ way to place the $R$'s in the remaining two positions. That gives a total number of permutations of:
$$\require{cancel} \dbinom{11}{5}\dbinom{6}{2}\dbinom{4}{1}\dbinom{3}{1} = \dfrac{11!}{5!\cancel{6!}}\cdot \dfrac{\cancel{6!}}{2!\cancel{4!}} \cdot \dfrac{\cancel{4!}}{1!\cancel{3!}}\cdot \dfrac{\cancel{3!}}{1!2!} = \dfrac{11!}{5!2!1!1!2!}$$
This method can be generalized for multiset permutations. Would you want to see a generalization, as well?
A: For the example AABBB:
If the letters were distinct (say, $A_1, A_2, B_1, B_2, B_3$), there would be $5!$ permutations. However, if you want the two $A$s to be indistinguishable, then you've counted words like $A B_2 B_1 A B_3$ twice: once as $A_1 B_2 B_1 A_2 B_3$ and once like $A_2 B_2 B_1 A_1 B_3$. To account for this, you divide by $2!$.
Similarly, if the $B$s are indistinguishable, you have counted words like $ABBAB$ six times: $AB_1 B_2 A B_3$, $A B_1 B_3 A B_2$, $A B_2 B_1 A B_3$, $A B_2 B_3 A B_1$, $A B_3 B_1 A B_2$, $A B_3 B_2 A B_1$. To account for this, you divide by $3!$.
A: Welcome to MSE. For a second, lets say that you paint your letters of different color. Say $\color{red}{A}B\color{blue}{B}\color{green}{A}$ then the letters with the same symbol now look different because of their color. No? In how many ways can you arrange them? $4!$ seems the right choice because you can distinguish between every letter and the number of orderings in $n$ distinct objects is $n!$ Now, erase the colors, notice that for a particular arrangement, you flip the colors in every possible way! So for an arrangement like $BAAB$ you actually had $B\color{red}{A}\color{green}{A}\color{blue}{B},B\color{green}{A}\color{red}{A}\color{blue}{B},\color{blue}{B}\color{red}{A}\color{green}{A}B$ and $\color{blue}{B}\color{green}{A}\color{red}{A}B.$ Notice then that if you fix a symbol you did the number of distinct letter different arrangements. By the multiplication principle, you did the product of the factorials of each individual symbol. So you have to divide by the total arrangements to get $1$ per group of those. That is, then how you get $\frac{4!}{\underbrace{2!}_{B}\cdot \underbrace{2!}_{A}}.$
