Probability of chosing at least two given letters out of three in sequence of five. I am self studying elementary Probability Theory, and I am not sure on how to solve this exercise, taken from the book Understanding Probability by Henk Tijms.

For the upcoming drawing of the Bingo Lottery, five extra prizes have
  been added to the pot. Each prize consists of an all-expenses paid
  vacation trip. Each prize winner may choose from among three possible
  destinations $A$, $B$ and $C$. The three destinations are equally
  popular. The prize winners choose their destination indipendently of
  each other. Calculate the probabilty that at least one of the
  destinations $A$ and $B$ will be chosen. Also, calculate the
  probability that not each of the three destinations will be chosen.

Regarding the first question, at first I thought that the probability that at least one of the destinations $A$ and $B$ is chosen should be the complementary probability that all the five winners choose the destination $C$, so:
$$P(A\cup B)=1-P(C)=1-1/3^5=\frac{242}{243}.$$
I would like to compute the same probability through the inclusion-exclusion rule,
$$P(A\cup B) = P(A)+P(B)-P(A\cap B).$$
In this case it should be $P(A)=P(B)$, and to compute $P(A)$ I consider first the sequences of five letters containing exactly one time the letter $A$, which are  $\binom{5}{1}$ and have probability $\frac13 (\frac23)^4$.
Similarly, the sequences of five letters containing exactly two $A$s are $\binom{5}{2}$ and have probability $(\frac13)^2(\frac23)^3$. In this way, I get
$$P(A)=\sum_{k=1}^5 \binom{5}{k}\left(\frac13\right)^k\left(\frac23\right)^{5-k}.$$
To compute the probability that both the destinations $A$ and $B$ are chosen, I start by computing the probability that $A$ and $B$ are chosen one time each, and the remaining three winners choose $C$. In this case we have $\binom{5}{2}$ ways of forming the string $ABCCC$, which has probability $1/3^5$. Next, consider the two sequences $AABCC$ and $ABBCC$. The number of sequences are $2\cdot\binom{5}{3}$ and the probability is again $1/3^5$.
Summing up, I have
$$P(A\cap B)=\frac{1}{3^5}\sum_{k=2}^5 \binom{5}{k}(k-1).$$
However, when I compute the numerical values for $P(A\cup B)$, I get a number greater than one so I should be making a mistake somewhere.
 A: For your calculation of $P(A\cap B)$, it appears you're missing some cases.

Here's one way to organize the count . . .
$$
\begin{array}
{|c|c|c|c|} 
A&B&C&\text{count}&\text{times}\\ \hline
1&4&0&{\large{\binom{5}{1}}}{\large{\binom{4}{4}}}=5&2\\
2&3&0&{\large{\binom{5}{2}}}{\large{\binom{3}{3}}}=10&2\\ \hline
1&3&1&{\large{\binom{5}{1}}}{\large{\binom{4}{3}}}{\large{\binom{1}{1}}}=20&2\\
2&2&1&{\large{\binom{5}{2}}}{\large{\binom{3}{2}}}{\large{\binom{1}{1}}}=30&1\\ \hline
1&2&2&{\large{\binom{5}{1}}}{\large{\binom{4}{2}}}{\large{\binom{2}{2}}}=30&2\\ \hline
1&1&3&{\large{\binom{5}{1}}}{\large{\binom{4}{1}}}{\large{\binom{3}{3}}}=20&1\\ \hline
\end{array}
$$
The entries in the "times" column are symmetry correction factors. For example, the count for the triple $(A,B,C)=(1,4,0)$ needs to be multiplied by $2$, so as to include the count for the symmetrically equivalent triple $(A,B,C)=(4,1,0)$.

Thus, the total count is
$$(5)(2)+(10)(2)+(20)(2)+(30)(1)+(30)(2)+(20)(1) =180$$
hence 
$$P(A\cap B) = \frac{180}{3^5} = \frac{20}{27}$$
so then
\begin{align*}
P(A\cup B) &=P(A)+P(B) - P(A\cap B)\\[6pt]
&=2\left(\frac{211}{3^5}\right)- \frac{20}{27}\\[6pt]
&=\frac{242}{243}
\end{align*}
as expected.

For the second problem, first find the probability that all $3$ destinations are chosen.

As shown in the table below, there are only two cases . . .
$$
\begin{array}
{|c|c|c|c|} 
A&B&C&\text{count}&\text{times}\\ \hline
1&1&3&{\large{\binom{5}{1}}}{\large{\binom{4}{1}}}{\large{\binom{3}{3}}}=20&3\\ \hline
1&2&2&{\large{\binom{5}{1}}}{\large{\binom{4}{2}}}{\large{\binom{2}{2}}}=30&3\\ \hline
\end{array}
$$
Thus, the total count is
$$(20)(3)+(30)(3)=150$$
hence 
$$P(A\cap B \cap C) = \frac{150}{3^5} = \frac{50}{81}$$
so the probability that not all $3$ destinations are chosen is
$$1- \frac{50}{81} = \frac{31}{81}$$
