How many different 4 letter words can you create from the characters of parallelogram?
I have absolutely no idea where to start on this one. Can someone please point me in the correct direction?
Thanks!
EDIT:
This is what I have so far
$$\begin{align}
&\text{Case 1: }\binom{2}{1}*\binom{4}{3}*\binom{7}{3} = 280 \\
&\text{Case 2: }\binom{3}{2}*\binom{4}{2} = 18\\
&\text{Case 3: }\binom{3}{1}*\binom{7}{2} = 63\\
&\text{Case 4: }\binom{8}{4} = 70\\
\end{align}$$
So the answer I get is $431$. Would this be correct?
Best Answer
The following is a little ugly. Note that we have $3$ l's, $3$ a's, 2 r's, and $5$ singleton letters.
Divide into cases:
(i) If we will use $3$ l's, their locations can be chosen in $\binom{4}{1}$ ways, and the remaining letter can be chosen in $\binom{7}{1}$ ways, for a total of $\binom{4}{1}\binom{7}{1}$.
There are also $\binom{4}{1}\binom{7}{1}$ words with $3$ a's.
Or else we could have said the tripled letter can be chosen in $\binom{2}{1}$ ways. for each choice, where it goes can be chosen in $\binom{4}{3}$ ways, and then the remaining slot can be filled in $\dots$.
(ii) If the word will have $2$ doubled letters, the types of the letters can be chosen in $\binom{3}{2}$ ways. For each of these ways, the places where the earlier (in the alphabet) of the chosen letters go can be chosen in $\binom{4}{2}$ ways.
(iii) If the word will have exactly $1$ doubled letter, that letter can be chosen in $\binom{3}{1}$ ways. Continue.
(iv) If the word is to have no repeated letter, we have a simple situation, an alphabet of $8$ letters.