2015 Mathcounts National sprint #22 level 1 +

 

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 \[ D_3 \;=\; 3!\left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!}\right) \;=\; 6\left(1 - 1 + \frac{1}{2} - \frac{1}{6}\right) \;=\; 2. \]