1

How many permutations are there in the word "arrangement" that have exactly two of the following consecutive letters in them: aa,ee,rr,nn? For example: "aarrngement"

N. F. Taussig
  • 76,571
Ehsan
  • 25
  • 5

1 Answers1

3

Inclusion-exclusion can be applied, but it is a bit tricky. not exactly run-of-the-mill.

ways with at least $2$ together - ways with at least $3$ together + ways with all $4$ together

= $\binom42\binom22\frac{9!}{2!2!} - \binom43\binom32\frac{8!}{2!} + \binom44\binom427! = 332,640$

Try to work out the logic for the extra coefficients..

Added, a way without inclusion-exclusion:

Let us take $AA, EE$ together as $\mathcal A, \mathcal E$, respectively.

Let us then separate the $RR$ by inserting in gaps $\uparrow \mathcal{ \large A}\uparrow \mathcal{\large E}\uparrow N\uparrow N\uparrow G\uparrow M\uparrow T\uparrow$ and permute the rest. This can be done in $\binom82\times\frac{7!}{2!}$ ways.

Now, from such arrangements, subtract those with $NN$ together,
$\uparrow \mathcal{ \large A}\uparrow \mathcal{\large E}\uparrow\mathcal{\large N}\uparrow G\uparrow M\uparrow T\uparrow$ in $\binom72\times 6!$ ways

$\binom82 \times \frac{7!}{2!} - \binom72 \times 6! = 55440$

And since there are $\binom42 = 6$ ways of choosing the initial "together pair", final answer is $6\times 55440 = 332,640$, as before.

true blue anil
  • 41,295
  • You provided two nice solutions to a tricky problem. However, there is a typographical error you should fix. The $21$ in your calculation of the $55440$ arrangements in which AA and EE are the only double letters should be $2!$. – N. F. Taussig Mar 19 '16 at 10:55
  • @N.F.Taussig: Thanks, corrected. – true blue anil Mar 19 '16 at 18:04