How many permutations of 11 elements are there, from a set of 12 elements with $3$ a's, $3$ b's, $3$ c's, and $3$ d's?
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.
Solution 1:
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:
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Everything except 1 $a$: $\{a,a,b,b,b,c,c,c,d,d,d\}$
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Everything except 1 $b$: $\{a,a,a,b,b,c,c,c,d,d,d\}$
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Everything except 1 $c$: $\{a,a,a,b,b,b,c,c,d,d,d\}$
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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.
Solution 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:
$\hspace{.3 in}\displaystyle \binom{12}{3}\binom{9}{3}\binom{6}{3}=\frac{12!}{(3!)^4}$