2015 Mathcounts National
sprint #22
If six
people randomly sit down at a table with six chairs, and they do not notice
that there are name tags marking assigned seats, what is the probability that
exactly three of them sit in the seat he or she was assigned? Express your
answer as a common fraction.
Try this yourself first (extremely), then scroll down for solutions.
Method 1-Elementary Counting
Choose the three people who sit correctly:
\( \binom{6}{3}=20 \).
The remaining three must all sit incorrectly.
A quick check (or listing) shows only two ways: \((A\,B\,C)\) or \((A\,C\,B)\).
Total favourable seatings \(=20\times2=40\); total seatings \(=6!\).
Therefore \( P=\dfrac{40}{720}=\boxed{\tfrac{1}{18}} \).
Method 2 – Derangement formula
Pick the three fixed seats: \( \binom{6}{3}=20 \).
Derange the other three: \( !3 = 2 \).
Again \( 20\times2 = 40 \) good seatings, so
\[
P \;=\; \frac{20\cdot!3}{6!} \;=\; \boxed{\tfrac{1}{18}} .
\]
Derange the remaining 3 people. The number of derangements of 3 items is \[ D_3 \;=\; 3!\left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!}\right) \;=\; 6\left(1 - 1 + \frac{1}{2} - \frac{1}{6}\right) \;=\; 2. \]
