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From the letters "a,b,c" we assemble words with $7$ letters, how many words can we assemble if each letter should appear at least twice.

My attempt:

There are $7$ places: $\color{blue}{*******}$

$1)\color{red}{a}\color{blue}{b}\color{green}{c}\color{red}{a}\color{blue}{b}\color{green}{c}\color{red}{a}$

$2)\color{red}{a}\color{blue}{b}\color{green}{c}\color{red}{a}\color{blue}{b}\color{green}{c}\color{blue}{b}$

$3)\color{red}{a}\color{blue}{b}\color{green}{c}\color{red}{a}\color{blue}{b}\color{green}{c}\color{green}{c}$

$\Longrightarrow\boxed{ 3\cdot \binom{7}{3}\cdot \binom{4}{2}\cdot \binom{2}{2}}$

English is not my first language, so feel free to correct

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    $\begingroup$ The answer $3\cdot\binom{7}{3}\cdot\binom{4}{2}$ looks fine to me. $\endgroup$ – Jack D'Aurizio Oct 25 '15 at 12:48
  • $\begingroup$ @JackD'Aurizio ok thanks $\endgroup$ – 3SAT Oct 25 '15 at 12:51
  • $\begingroup$ The answer is ok, but the explanation simply isn't there. For all I know, you could have peeked your solution from someone else, so I would still give 0 marks for this. $\endgroup$ – Jacopo Notarstefano Oct 25 '15 at 12:59
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The answer is ok, but you might use another approach if you like it.

Choose the extra letter, and permute all $7$, so $\left(3\cdot\dfrac{7!}{3!2!!2!}\right)$

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