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.

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 
$$|U| - |A \cup B \cup C| = 5^7 - \binom{3}{1}4^7 + \binom{3}{2}3^7 - \binom{3}{3}2^7$$ 
