<s>[INST] Find the number of seven-letter words that use letters from the set $\{\alpha,\beta,\gamma,\delta, \epsilon\}$... Find the number of seven-letter words that use letters from the set $\{\alpha,\beta,\gamma,\delta, \epsilon\}$ and contain at least one each of $\alpha$, $\beta$, and $\gamma$.

My attempt: using inclusion/exclusion 

Let A denotes $\alpha$ is missing, B denotes $\beta$ is missing, and C denotes $\gamma$ is missing. Then, $$\begin{aligned}|A\cup B\cup C|&=|A|+|B|+|C|-(|AB|+|AC|+|BC|)+|ABC|\\ &= 2^7+2^7+2^7-(1+1+1)-0\\ &= 381\end{aligned}$$
    $|U|= 3^7 (\text{No restriction on the combination of words})$ 
So, $$\begin{aligned} |U|-|A\cup B\cup C|&= 3^7-[2^7+2^7+2^7-(1+1+1)-0]\\ &= 1806 \end{aligned}$$


I'm not sure if I'm doing this problem correctly at all, because I think the universal set, i.e $|U|$, should be $5^7$ and I not sure if my calculation needs to include the other elements as well. I'm just second guessing myself. Please help me. Thank you.
 [/INST] Since there are five letters available and each word has seven positions to fill, there are $5^7$ words.  From these, we must exclude those cases in which one or more of the letters $\alpha$, $\beta$, or $\gamma$ is missing.
There are three ways to exclude one of the letters $\alpha$, $\beta$, or $\gamma$ and four ways to fill each of the seven positions with the other four letters.  Hence, there are 
$$\binom{3}{1}4^7$$
ways to exclude one of the letters $\alpha$, $\beta$, $\gamma$.
There are three ways to exclude two of $\alpha$, $\beta$, or $\gamma$ and three ways to fill each of the seven positions with the other three letters.  Hence, there are 
$$\binom{3}{2}3^7$$
ways to exclude two of the letters $\alpha$, $\beta$, $\gamma$.  
There is one ways to exclude all three of the letters $\alpha$, $\beta$, and $\gamma$ and two ways to fill each of the seven positions with the other two letters.  Hence, there are 
$$\binom{3}{3}2^7$$
ways to exclude all three of the letters $\alpha$, $\beta$, and $\gamma$.  
By the Inclusion-Exclusion Principle, the number of seven-letter words that can be formed from the set $\{\alpha, \beta, \gamma, \delta, \epsilon\}$ that contain at least one $\alpha$, at least one $\beta$, and at least one $\gamma$ is 
$$5^7 - \binom{3}{1}4^7 + \binom{3}{2}3^7 - \binom{3}{3}1^7$$
If we let $U$, $A$, $B$, and $C$ be the sets you named, then 
\begin{align*}
|U| & = 5^7\\
|A| & = 4^7\\
|B| & = 4^7\\
|C| & = 4^7\\
|A \cap B| & = 3^7\\
|A \cap C| & = 3^7\\
|B \cap C| & = 3^7\\
|A \cap B \cap C| & = 2^7
\end{align*}
Hence, 
\begin{align*}
|A \cup B \cup C| & = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|\\ 
                  & = 4^7 + 4^7 + 4^7 - 3^7 - 3^7 - 3^7 + 2^7\\
                  & = 3 \cdot 4^7 - 3 \cdot 3^7 + 2^7\\
                  & = \binom{3}{1}4^7 - \binom{3}{2} 3^7 + \binom{3}{3}2^7
\end{align*}
Thus, the number of permissible words is 
$$|U| - |A \cup B \cup C| = 5^7 - \binom{3}{1}4^7 + \binom{3}{2}3^7 - \binom{3}{3}2^7$$ 
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