Same Arrangements of the word "MINIMUM" In how many distinguishable ways can the seven letters in the word MINIMUM be arranged, if all the letters are used each time?
My attempt:
3!(2!) = 12 ways.
M has 3 choices and I has two choices. These ways are mutually exclusive(independent) so multiply.
How am I wrong? 
 A: Since MINIMUM has seven letters, you have seven positions to fill.  You can fill three of them with M's in $\binom{7}{3}$ ways.  You now have four positions to fill.  You can fill two of them with I's in $\binom{4}{2}$ ways.  You now have two positions left to fill.  You can fill one of them with an N in $\binom{2}{1}$ ways.  You can fill the final position with a U in $\binom{1}{1}$ way.  Hence, the number of distinguishable arrangements of the letters of the word MINIMUM is 
$$\binom{7}{3}\binom{4}{2}\binom{2}{1}\binom{1}{1} = \frac{7!}{4!3!} \cdot \frac{4!}{2!2!} \cdot \frac{2!}{1!1!} \cdot \frac{1!}{1!0!} = \frac{7!}{3!2!1!1!}$$ 
A: This can be found using the multinomial theorem.  Here, we have 
$$\frac {7!}{3!2!}\,\text{ possible distinct combinations}$$ where $7$ is the the length of the word "minimum", $3$ is the number of "m's" we have to work with, and $2$ is the number of "i"'s we have on hand to work with.  
A: Pretending that you have a 7 letter word with all different letters, there would be 7!. However since there are repeating letters, it's 7!/(3!2!1!1!) because there's 3 M's, 2 I's, 1 N, and 1 U. 
For more reading, see this link. Read under Distinguishable Permutations.
