2

How many distinct 4-letter arrangements can be made with the letters in the word "PARALLEL"?

My approach: Because we are only looking at how many different permutations there are and not the frequency at which these permutations exist, we can delete the repeated letters and leave only one. This leaves us with the following set of letters: $\{P, A, R, L, E\}$ So $5$ permute $4$ is $120$. It was only then that I realized that deleting repeats will remove words such as $LLLE$ to exist. At this point, I do not know how to add on these possibilities to my approach.

I would appreciate help, and as always, bash me whenever you see a typical blunder of mine.

Ian L
  • 899

4 Answers4

4

As you can see, the case approach can be quite error prone, especially for more complex problems:

An alternative approach using generating functions is to find:

coefficient of $x^4$ in $4!(1+x)^3(1+x+\frac{x^2}{2!})(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}) = 286$

For a detailed explanation of why this works, you can see this answer on MSE

Community
  • 1
true blue anil
  • 48,250
  • The only "right" approach, IMO. – Caleb Stanford Aug 10 '16 at 04:28
  • @trueblueanil The problem though is how do you actually evaluate a generating function like that? With all those factorials on the denominators of the $x$-terms, it doesn't follow the usual identities for manipulating generating functions. At some point how is this any different than doing it the long way? – AJJ Aug 10 '16 at 14:47
  • @ArukaJ: If you aren't having computational help, it is not so good ! But it is more mechanical. If you fill up the $4$ slots with the $3$ types, you will find $10$ patterns, $310,301,220,211,202,121,112,103,022,013$, and you should be able to figure out that they will yield $24[1+1+3(\frac12+1+\frac12+\frac12+\frac12+\frac16)+\frac14+\frac16)] = 286$ – true blue anil Aug 10 '16 at 15:58
  • Fill up the slots with types? What do you mean? Don't you have to multiply out that huge polynomial? – AJJ Aug 10 '16 at 16:46
  • @Aruka J: In lieu of multiplying out ! The $3$ types are the letters that are singles $(P,R,E)$, doubles $(A's)$ and triples $(L's). 121$, e.g. represents one from singles, two from doubles and one from triple. When choosing from singles, you only need to remember to use the appropriate binomial coefficient $\left[\binom31; here\right],$ and for "multiples" you have to remember to divide by the appropriate factorial $\left[2!; here\right]$. All said and done, it is still a painful process, but much more mechanical. – true blue anil Aug 10 '16 at 17:02
2

There are a few cases:

  1. All $4$ letters distinct. There are $\binom{5}{4} = 5$ ways to pick the letters and $4!$ ways to order them.

  2. Two letters distinct, one letter repeated twice. Either two A's or two L's. If we have two A's, then we have $\binom{4}{2} = 6$ ways to pick the other letters. $4!$ ways to order them, but divide by $2!$ to account for the doubled letter. Then do the same thing again for the double-L case.

  3. Two letters repeated twice. Two A's, two L's. $4!$ ways to order the letters, divide by $2!2!$ to account for the repetitions.

  4. One letter by itself, one letter repeated three times. Just the L here. Once we have the triple L, there are $4$ choices for the final letter. $4!$ ways to arrange the letters, divided by $3!$ for the triple L.

$$5 \cdot 4! + \frac{6 \cdot 4! \cdot 2}{2!} + \frac{4!}{2! \cdot 2!} + \frac{4 \cdot 4!}{3!} = 286$$

AJJ
  • 2,083
  • 3
  • 18
  • 30
1

Arranging the letters in alphabetical order yields AA E LLL P R. In the four-letter word there may be

  • no repeats: this has been calculated in the question, and there are 120 words of this form.
  • two of A, the rest distinct: There are $\binom42=6$ ways to place the A's, and 4 × 3 = 12 ways to place two more letters in the word. This makes 72 ways for this case.
  • two of L, the rest distinct: this has the same number of words as the previous case (72).
  • two of A and two of L: Again, 6 ways to place A's, but then there is only one way to place the L's. Hence 6 ways.
  • three of L: there are $\binom43=4$ ways to place the L's, and 4 possibilities for the remaining letter. This makes 16 ways.

Summing all cases up (which are mutually exclusive and exhaustive) yields 286 4-letter words from PARALLEL.

Parcly Taxel
  • 105,904
0

If you use $j$ A's and $k$ L's, there are $\binom3{4-j-k}$ ways to select the remaining letters, and there are $\frac{4!}{j!k!}$ different ways to arrange the $4$ letters, so in total there are

$$ \sum_{j=0}^2\sum_{k=0}^3\binom3{4-j-k}\frac{4!}{j!k!}=286 $$

admissible words (Wolfram|Alpha computation of the double sum).

joriki
  • 242,601