Problem 23 - A First Course in Probability

Question

A psychology laboratory conducting dream research contains 3 rooms, with 2 beds in each room. If 3 sets of identical twins are to be assigned to these 6 beds so that each set of twins sleeps in different beds in the same room, how many assignments are possible?

Step-by-Step Solution

Answer

There are 8 possible assignments for the 3 sets of identical twins to sleep on different beds in the same room.

Step 1: Arranging Twins In Rooms

First, we need to find out how many ways we can assign the twins to the rooms. Since there are 3 sets of twins and 3 rooms, this becomes a straightforward combinatorics problem. We need to choose 3 rooms for the 3 sets of twins, so we do this using the combination formula: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] In this case, \(n = 3\) (the number of rooms) and \(k = 3\) (the number of sets of twins), so we have: \[ \binom{3}{3} = \frac{3!}{3! (3-3)!} = 1 \] There is only 1 way to assign the twins to the rooms since there are the same number of sets of twins as there are rooms.

Step 2: Arranging Twins In Beds

Now that the twins are assigned to the rooms, we need to assign each twin to a bed within their assigned room. Since there are 2 beds in each room, there are 2 ways to assign the twins to beds for each set.

Step 3: Calculating Total Assignments

To find the total number of possible assignments, we simply multiply the number of arrangements found in Step 1 and the number of arrangements found in Step 2. Since there are 3 sets of twins, and there are 2 ways to assign beds for each set, we have: \[ 1 * 2^3 = 1 * 8 = 8 \] There are 8 possible assignments for the 3 sets of identical twins to sleep on different beds in the same room.

Key Concepts

Probability Theory

When we discuss probability theory, we refer to a branch of mathematics that deals with the likelihood of different outcomes in random events. This includes analyzing various scenarios to understand how probable certain outcomes are. In our exercise, the context isn't based on randomness, but on deliberate decisions. Even so, the foundations of probability theory guide us in quantifying the number of possible ways to arrange objects — in this case, the assignment of twins to beds.

Understanding the probability of certain arrangements can extend to evaluating more complex situations where the likelihood of events is not equally distributed, such as the chance of rolling a six on a loaded dice versus a fair dice. In the given problem, since there is a finite number of twins and beds, and each twin must occupy a unique bed, we actually explore a deterministic combinatorial scenario rather than a probabilistic one. However, probability theory principles provide a useful framework for solving this kind of combinatorial problem.

Combinatorial Analysis

In our exercise, combinatorial analysis is pivotal. This area of mathematics focuses on counting, arrangement, and combination of sets of elements. The heart of combinatorial analysis is understanding how to systematically count without having to list each possibility, which becomes practically impossible with larger sets.

For instance, when assigning twins to beds, we must consider each unique arrangement without repetition. This form of counting is fundamental in combinatorial analysis. It includes the use of permutation for ordered arrangements and combinations where the order doesn't matter. Here, we're dealing with combinations because the order, which twin gets which bed, isn't important. Thus, the combination formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \) is used, encapsulating the essence of combinatorial analysis by calculating how many ways we can choose k items from a set of n without regard to the order.

Factorial Calculation

Digging deeper into the exercise, the concept of factorial calculation is fundamental. The factorial of a non-negative integer n, denoted by n!, is the product of all positive integers less than or equal to n. It represents the number of ways to arrange n distinct items into a sequence, known as a permutation.

The factorial function grows at a very fast rate with an increase in n, which makes the exploration of certain combinatorial problems challenging without the use of this mathematical shorthand. For example, in the combination formula used for our exercise \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), factorials are used both in the numerator and the denominator to simplify the process of counting combinations without repetition. Understanding how to compute and manipulate factorials is crucial in solving many combinatorial problems.