Question

5 students, all with distinct ages, are randomly seated in a row at the movies. The probability that, from left to right, no three consecutive students are seated in increasing age order is $\frac{m}{n}$, where m and n are relatively prime positive integers. Find m + n.

Answer 1

19

Answer 2

Here's how to solve this probability problem:

1. Total arrangements: There are $5! = 120$ ways to seat 5 students.

2. Counting "bad" arrangements (at least one increasing triple): Let the positions be 1–5 and define:

   • $A_1$: positions 1,2,3 in increasing order.
   • $A_2$: positions 2,3,4 in increasing order.
   • $A_3$: positions 3,4,5 in increasing order.

3. Individual counts: For any triple (say $A_1$):

   • Choose any 3 out of 5 students: $\binom{5}{3}=10$.
   • They can be arranged in exactly 1 increasing order.
   • Arrange the remaining 2 in $2!$ ways.
   • So, $|A_1| = 10 \times 2 = 20$. Similarly, $|A_2| = |A_3| = 20$.

4. Pairwise intersections:

   • $A_1 \cap A_2$: This forces positions 1–4 to be in increasing order. Choose 4 out of 5: $\binom{5}{4}=5$ ways, and the remaining one is fixed. Thus, 5 ways.
   • $A_2 \cap A_3$: Forces positions 2–5 to be increasing. Similarly, 5 ways.
   • $A_1 \cap A_3$: Here, positions 1,2,3 must be increasing and positions 3,4,5 increasing. The common position (seat 3) forces the overall condition that the number in position 3 acts as a "pivot". For 5 distinct numbers, the only possibility is when the number in seat 3 has exactly two numbers smaller (in seats 1 and 2) and two larger (in seats 4 and 5). Only $3$ can be the pivot (with $1,2$ before and $4,5$ after). So, there is exactly 1 way.

5. Triple intersection: $A_1 \cap A_2 \cap A_3$ forces the entire row to be in increasing order. (1 way)

6. Using Inclusion-Exclusion:

   $$|\text{At least one increasing triple}| = |A_1|+|A_2|+|A_3| - (|A_1\cap A_2|+|A_2\cap A_3|+|A_1\cap A_3|) + |A_1\cap A_2\cap A_3|$$ $$= 20+20+20 - (5+5+1) + 1 = 60 - 11 + 1 = 50.$$

7. Favorable arrangements: Arrangements with no three consecutive in increasing order:

   $$120 - 50 = 70.$$

   The required probability is $\frac{70}{120} = \frac{7}{12}$, so $m=7$ and $n=12$. Therefore,

   $$m+n=7+12=19.$$

Core Explanation:

• Total arrangements = $120$.
• Count "bad" cases with at least one increasing triple using Inclusion-Exclusion = $50$.
• Good arrangements = $120-50=70$.
• Probability = $\frac{70}{120}=\frac{7}{12}$ so $m+n=19$.