1

Let $S=\{a,a,a,b,b,b,c,c,c,d,d,d\}$. Determine the number of $11$-permutations of the set $S$.

What's the number of ways to take 11 objects from this set S with 12 objects, such that the order of the objects does matter?

I'm not sure if $P(12,11)/(3!)^4$ is the correct approach or not.

TAPLON
  • 1,956
  • 1
  • 19
  • 40
  • What is an $11$ permutation? – Arthur Sep 03 '16 at 22:23
  • The number of ways to take 11 objects from the set S such that the order does not matter. – TAPLON Sep 03 '16 at 22:25
  • 1
    @Arthur I would interpret it as "How many length 11 strings exist using 0-3 a's, 0-3 b's, 0-3 c's and 0-3 d's." An interesting thing to note about this specific situation is that you can find a nice bijection to an easier problem: that of the one where you actually arrange all twelve characters and cross through the last appearing one. – JMoravitz Sep 03 '16 at 22:25
  • 2
    @JustinStevenson you mean where order does matter. Notice further that $P(12,11) = \frac{12!}{(12-11)!}=\frac{12!}{1!}=12!$, so your proposed answer is in fact equal to $\binom{12}{3,3,3,3}$ and is correct, but probably for the wrong reasons. If you were to try to answer the question of how many 2-permutations following your proposed formula, I expect you will get a number which is not even an integer. – JMoravitz Sep 03 '16 at 22:27
  • Ah yes I did mean where order does matter, and if using P(12,11) isn't the correct approach what would be the correct approach? – TAPLON Sep 03 '16 at 22:33
  • 1
    Brute force can handle most small cases and is probably the easiest approach for small cases. By brute force, I mean, break into cases for what the unused letters are. For larger problems, it becomes a very difficult problem., For example, "How many 100 permutations of {30a, 30b, 30c, 30d}" I would not want to do. Read this question for the start of more general techniques. – JMoravitz Sep 03 '16 at 22:36

2 Answers2

3

You got the right answer, but for the wrong reason :)

The set only has $11$ objects, so in order to take $11$ of them, we have to just choose $1$ to not take. So here are the possibilities for the $11$ objects we can take:

  • Everything except 1 $a$: $\{a,a,b,b,b,c,c,c,d,d,d\}$

  • Everything except 1 $b$: $\{a,a,a,b,b,c,c,c,d,d,d\}$

  • Everything except 1 $c$: $\{a,a,a,b,b,b,c,c,d,d,d\}$

  • Everything except 1 $d$: $\{a,a,a,b,b,b,c,c,c,d,d\}$

In each of the 4 cases, the number of ways of arranging the 11 objects will be $$ \frac{11!}{3!3!3!2!}. $$ Note that the 2! is not a typo: one of a, b, c, d only appears 2 times (rather than 3) in a set of 11. Therefore, since there are $4$ cases in the list above, the total number of possibilities is $$ 4 \cdot \frac{11!}{3!3!3!2!} = \underbrace{\frac{12!}{(3!)^4}}_{\text{Your answer}} = 369600. $$

Quicker way

A quicker way is described in the comments and other answer: you can first notice that the number of ways to list out 11 objects is the same as the number of ways to list out 12. Then you are just counting the number of ways to arrange 12 objects where 4 sets of 3 are repeated.

Caleb Stanford
  • 47,093
  • Is 2! a typo in your last line? Shouldn't 2! be 3!? Because a, b, c, d each exist thrice in S? –  Jan 03 '22 at 07:26
  • @PGTK It's not a typo: one of a, b, c, d only exists twice in a set of 11 elements of S. I have updated the answer to clarify that. – Caleb Stanford Jan 03 '22 at 07:37
  • thanks for updating your answer! can you please clarify the 4 at the front of $4 \cdot \frac{11!}{3!3!3!2!}$? Why multiply by 4? –  Jan 06 '22 at 07:32
  • Pls edit your answer, rather than comment? Comments can be deleted. –  Jan 06 '22 at 07:32
2

Every 11-permutation of S corresponds uniquely to a permutation of S, by inserting the missing element of the 11-permutation at the end.

(For example, $\;\;acbdacbddab\longleftrightarrow acbdacbddabc$)

Therefore the number of 11-permutations of S is the same as the number of permutations of S:

$\hspace{.3 in}\displaystyle \binom{12}{3}\binom{9}{3}\binom{6}{3}=\frac{12!}{(3!)^4}$

user84413
  • 27,747