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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?



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?

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I assume this is a maths question, so you don't require proper English words? (For example is prlg a valid word?) – Tara B Feb 28 '13 at 0:11
I can think of glop, lame, mare, leap, real, aloe, pall, goal, lore, oral, pear, pore, roll, gram, and pram. If proper names are allowed, there's lear and lego :) – Trevor Wilson Feb 28 '13 at 0:15
I'm not sure how we feel about larp. – Trevor Wilson Feb 28 '13 at 0:16
@TaraB, no there is no need for proper English words :) Otherwise that would require more than just Math :) – Jeel Shah Feb 28 '13 at 0:25
Yes, I was pretty sure that would be the case. I just thought it would be good to clarify. – Tara B Feb 28 '13 at 0:27
up vote 2 down vote accepted

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.

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I get an answer of 431, would this be correct? – Jeel Shah Feb 28 '13 at 0:51
A lot more. For example, for Case (iv), there are $(8)(7)(6)(5)=1680$. For Case (iii) there are I think $756$. I gave explicit formulas for Cases (i) and (ii). We need to add the numbers obtained in the $4$ cases. – André Nicolas Feb 28 '13 at 1:03

There are $13$ letters in parallelogram. If they were all different, the number of $4$-letter "words" would be $13\times12\times11\times10$ --- can you see why?

But of course the $13$ letters are not all different. So you have to look at all the words that use the letter "a" once, and treat them specially; also, all the words that use "a" twice and three times, and similarly for "r", and for "l".

You asked for a start --- does this get you started?

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So, we do $13*12*11*10$ because of permutations? But how would you go about implementing the repetition portion? Could you provide more detail? – Jeel Shah Feb 28 '13 at 0:05
I thought you just asked for a start. Anyway, Andre has provided more detail. – Gerry Myerson Feb 28 '13 at 0:22

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