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?
Best Answer
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
$$\binom{7}{3}\binom{4}{2}\binom{2}{1}\binom{1}{1} = \frac{7!}{4!3!} \cdot \frac{4!}{2!2!} \cdot \frac{2!}{1!1!} \cdot \frac{1!}{1!0!} = \frac{7!}{3!2!1!1!}$$