all possible permutations of "aabcd"

Question

Is it the correct solution to the problem: in how many ways can we permute the string "aabcd" such that every permutation is only 5 characters long and only the set of ${a,b,c,d}$ characters can be used (each character has to be used). So "aabdc", "abbcd" are valid permutations. My solution is the following: $$4*\binom{5}{2}*3!$$ because we can place the duplicate character in the string in $\binom{5}{2}$ ways and then we just need to permute the remaining 3 characters. Lastly, there're 4 possible duplicate characters.

Answer 1

The number of ways to arrange 5 elements, two of which are identical is $\frac{5!}{2!}$. For example there are $60$ arrangements of $aabcd$. Then there are 4 different characters that can be the repeated character. So I believe the answer is $4*\frac{5!}{2!}=240$, which is the same answer you got using your method.

Answer 2

In response to the title question (as originally posted):

$$\text{ All possible permutations of "aabcd"}$$

This is an example in which multinomials will work.

$$\binom{5}{2,1,1,1} = \frac {5!}{2!1!1!1!}=\frac {5!}{2!} =\frac{120}2 = 60$$

You have a total of five letters to work with, in which the there are 2 of one letter, and only 1 of the remaining three letters available.

This strategy works well, for example, determining the number of different string existing, for example, when permuting the letters of **MISSISSIPPI"

EDIT AFTER OP's edit: There is a contradiction in the OP's problem statement. I answered the question in the title, when in fact the OP is not counting permutations.

If allowed to duplicate any one of a, b, c, d, to construct a string of five letters, with each of a, b, c, d must appear, then, since there are four distinct letters that could be duplicated, we simply multiply our previous answer by $4$ to get $$60\times 4 = 240$$ possible 5-letter strings, given these additional constraints.

Answer 3

There are $120$ distinct permutations of $abcde$. You can replace $e$ by any of $a',b',c',d'$, giving $480$ substitutes. Then drop the quote, resulting in every configuration represented twice. There are $240$ unique ones.

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Alternatively, you can scramble $aabcd$ in $5!$ ways, but each is repeated twice, hence $60$ distinct permutations. Same with $abbcd,abccd$ and $abcdd$.