Given the question that $6^3$=216 results of rolling a six-sided die three times, how many of them included getting exactly one 1, exactly one 2, or exactly one 6?
Here's my approach,
Let 'A' be the number of ways of getting exactly one '1' in rolling the die three times,
A = 3 x $5^2$ = 75 ways
Let 'B' be the number of ways of getting exactly one '2' in rolling the die three times,
B = 3 x $5^2$ = 75 ways
Let 'C' be the number of ways of getting exactly one '6' in rolling the die three times,
C = 3 x $5^2$ = 75 ways
The number of ways of getting exactly one '1' and exactly one '2' and exactly one '6' is
= 6 (Because the possible ways are {1,2,6}, {1,6,2}, {2,1,6}, {2,6,1}, {6,1,2}, {6,2,1})
A, B, and C are not disjoint. Therefore, by the principle of inclusion-exclusion, the number of ways of getting exactly one '1' or exactly one '2' or exactly one '6' should be
= (A + B + C) - (the number of ways of getting exactly one '1' and exactly one '2' and exactly one '6')
= 75 + 75 + 75 - 6
= 219
But the textbook answer is 159 which means that my approach is incorrect. The textbook doesn't include the explanation for the answer and hence, I'm requesting a help to find the mistakes/error in my solution.
Thanks in advanced.
You did not apply the Inclusion-Exclusion Principle correctly. For three sets, the Inclusion-Exclusion Principle states that $$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|$$
Let $A$ be the event that exactly one $1$ is obtained in three rolls of a die; let $B$ be the event that exactly one $2$ is obtained in three rolls of a die; let $C$ be the event that exactly one $6$ is obtained in three rolls of a die.
$|A|$: There are three ways to choose the position of the $1$. The remaining two spots must be filled by a number other than $1$. Hence, $$|A| = 3 \cdot 5^2$$ as you found.
By symmetry, $|A| = |B| = |C|$.
$|A \cap B|$: There are three ways to choose the position of the $1$, $2$ ways to choose one of the remaining positions for the $2$, and four possible numbers that could fill the remaining spot. Hence, $$|A \cap B| = 3 \cdot 2 \cdot 4$$
By symmetry, $|A \cap B| = |A \cap C| = |B \cap C|$.
$|A \cap B \cap C|$: There are $3!$ ways to arrange one $1$, one $2$, and one $6$, as you found.
Therefore, by the Inclusion-Exclusion Principle, the number of ways of obtaining exactly one $1$, exactly one $2$, or exactly one $6$ is $$|A \cup B \cup C| = 3 \cdot 3 \cdot 5^2 - 3 \cdot 3 \cdot 2 \cdot 4 + 3! = 159$$