Combinatorial Analysis: Seating Arrangements of Boys and Girls in a Row

In this article, we explore a combinatorial problem involving the seating arrangements of boys and girls in a row. Specifically, we want to determine the number of ways to seat 3 boys and 2 girls in a row such that each boy is seated next to at least one girl. We will use combinatorial principles to solve this problem, including the principle of inclusion and exclusion.

Introduction to the Problem

Suppose we have 3 boys named Alex, Bob, and Chris, and 2 girls. We are interested in arranging these 5 individuals in a row such that each boy is seated next to at least one girl. To solve this problem, we will define several sets of seating arrangements to facilitate our calculations.

Defining the Sets

We define the following sets:

A the set of seating arrangements where Alex is next to at least one girl. B the set of seating arrangements where Bob is next to at least one girl. C the set of seating arrangements where Chris is next to at least one girl.

The universal set (xi) represents all possible seating arrangements of the 5 individuals.

Calculating (midoverline{A} cup overline{B} cup overline{C}mid)

We use the principle of inclusion and exclusion to calculate the number of seating arrangements where at least one boy is not next to any girl. This is equivalent to finding the number of arrangements where none of the boys have a girl neighbor, and subtracting from the total number of arrangements.

Counting (midoverline{A}mid) (Complement of A)

(midoverline{A}mid) is the number of arrangements where Alex does not have a girl next to him. We break it down into cases:

Case 1: Alex has both Bob and Chris as neighbors on both sides. Case 2: Alex is at the left end, has Bob or Chris next to him. Case 3: Alex is at the right end, has Bob or Chris next to him.

Case 1: Alex and Neighbors

First, we place Alex in a fixed position and arrange Bob and Chris as his neighbors. This can be done in (2) ways (Bob-Christ- or Christ-Bob). The girls can occupy the remaining ends in (2) ways.

(midtext{Arrangements with Alex in the middle}mid 2 times 2 4)

Case 2: Alex at the Left End

When Alex is at the left end, he has Bob or Chris next to him. Bob can take the position to his right in (2) ways, and the remaining three can be arranged in (3! 6) ways.

(midtext{Arrangements with Alex at the left end}mid 2 times 6 12)

Case 3: Alex at the Right End

Similarly, when Alex is at the right end, the arrangement is a mirror of the previous case, so the number of arrangements is also (12).

Total ways 4 12 12 28)

By symmetry, (midoverline{B}mid midoverline{C}mid 28).

Counting (midoverline{A} cap overline{B}mid) and Others

Since the arrangements where Alex, Bob, and Chris do not have a girl neighbor are symmetric and mutually exclusive, we conclude that:

(midoverline{A} cap overline{B}mid midoverline{B} cap overline{C}mid midoverline{C} cap overline{A}mid midoverline{A} cap overline{B} cap overline{C}mid 0)

Final Calculation

Using the principle of inclusion-exclusion:

(mid overline{A} cup overline{B} cup overline{C} mid mid overline{A} mid mid overline{B} mid mid overline{C} mid - mid overline{A} cap overline{B} mid - mid overline{B} cap overline{C} mid - mid overline{C} cap overline{A} mid mid overline{A} cap overline{B} cap overline{C} mid)

( 28 28 28 - 0 - 0 - 0 0)

( 84)

By de Morgan’s law, we have:

(mid overline{A} cup overline{B} cup overline{C} mid mid overline{A cap B cap C} mid)

Therefore:

(mid A cap B cap C mid mid xi mid - mid overline{A} cup overline{B} cup overline{C} mid)

( 120 - 84 36)

Thus, the answer is boxed{36}.