How many ways can I list the letters?

Question

In how many ways can the letters $a, a, b, b, c, d, e$ be listed such that the letter $c$ and $d$ are not in consecutive positions?

My partial solution:

So, because we have $7$ letters, we will have $7!$ arrangements, but then we see that letters $a$ and $b$ have two copies, so we will have about $\frac{7!}{2!2!}$ arrangement, which according to my calculation is $1260$. But clearly, this number also counts all the arrangements where letters $c$ and $d$ are in consecutive order.

If I fix the letter $c$ in the first position, then I would have $5$ options for the second position, because I cannot have letter $d$ in the second position, and then $5!$ for the last five letters, so I will get $5*5!$. Same thing happens if I fix letter $d$ at the first position. But, then again they can be in other positions too. Plus, if letter $c$ is somewhere in the middle of the listing, then letter $d$ cannot be neither to its left nor to its right, which further limits the number of arrangements. I'm really stuck with it.

Answer 1

Take out $c$ and $d$ for the moment

Permute $aabbe$ in $\dfrac{5!}{2!2!}$ ways

Put back $c$ and $d$ in the gaps of permutations of $-a-a-b-b-e-$ in $6\cdot5$ ways

Putting everything together, $6\cdot5\dfrac{5!}{2!2!}$

Answer 2

I like the argument provided by true blue anil. To make your argument work, we must subtract the number of arrangements in which $c$ and $d$ appear consecutively from the total number of arrangements.

As you determined, the number of distinguishable arrangements of $a, a, b, b, c, d, e$ is $$\frac{7!}{2!2!}$$ To determine the number of arrangements in which $c$ and $d$ are consecutive, we treat $c$ and $d$ as a single unit. That gives us six objects to permute, of which two are $a$'s, two are $b$'s, one is the unit consisting of $c$ and $d$, and one is $e$. We can permute these objects in $\frac{6!}{2!2!}$ distinguishable ways and permute the unit consisting of $c$ and $d$ internally in $2!$ ways. Therefore, the number of arrangements in which $c$ and $d$ appear consecutively is $$2! \cdot \frac{6!}{2!2!} = \frac{6!}{2!}$$ Hence, the number of arrangements of the letters $a, a, b, b, c, d, e$ in which the letters $c$ and $d$ do not appear consecutively is $$\frac{7!}{2!2!} - \frac{6!}{2!}$$

Answer 3

$\frac{7!}{2! \times 2!}$ - The ways in which you can arrange the letters (including the cases when 'c' and 'd' are in consecutive positions).

$\frac{2! \times 5! \times 6}{2! \times 2!}$ - the cases when 'c' and 'd' are in consecutive positions.

$\frac{7!}{2! \times 2!} - \frac{2! \times 5! \times 6}{2! \times 2!} = \frac{7!}{4}-5! \times 3$ - this is what you want