Multiplication Rule, Factorials and Combinations

In your statistics class, you probably came across multiplication rule, factorials, and combinations, while solving probability questions. It's understandable that it can be confusing when to use which of these operations. Here's a brief overview along with examples to clarify their usage.

Multiplication Rule

Use the multiplication rule when you need to determine the total number of outcomes for a series of independent choices.

Example: Choosing 3 students from a classroom, with each student being either male (M) or female (F).

• Each choice has 2 possible outcomes (M or F).
• Total sequences = 2×2×2=8.

Example: Choosing a type of drink (coffee, tea, juice) and a type of pastry (croissant, muffin). Total combinations = 3×2 = 6.

Factorials

Use factorials when you need to count the number of ways to arrange a set of items where the order matters.

Example: Arranging 5 books on a shelf.

• The number of ways to arrange 5 books = 5! = 5×4×3×2×1=120.
  
  
• To understand this, imagine there are five slots on the shelf for the books. You have five choices for the first slot since any of the books can go there. After placing a book in the first slot, you have four remaining choices for the second slot, and so on.

Example: How many ways can 4 runners finish a race? Total arrangements = 4! = 24.

Combinations

Use combinations when you need to count the number of ways to choose items from a set without regard to the order.

Example: Choosing 3 students from a class of 10 to form a committee.

• The number of ways to choose 3 students from 10 = C(10, 3) = 10! / (3! x 7!) = 120

Example: Choosing 3 toppings from 5 available toppings for a pizza.

• The number of ways to choose 3 toppings from 5 = C(5, 3) = 5! / (3! x 2!) = 10

Summary

• Use the multiplication rule when you have independent choices.
• Use factorials when arranging items in a specific order.
• Use combinations when choosing items without regard to order.

 

Scenario: Planning a School Science Fair

Let's create a comprehensive scenario that involves a combination of the multiplication rule, factorials, and combinations to solve different aspects of the problem.

Imagine you are organizing a school science fair. You have to decide on several aspects:

1. Booth Assignment: Assign booths to different science projects. There are 3 science projects and 5 available booths. Each project can be assigned to any booth, but each project must have its own booth.
   
   
2. Volunteer Scheduling: Arrange volunteers to manage the event. There are 4 volunteers to manage different roles during the fair. Each volunteer will be assigned a specific role.
   
   
3. Selecting Judges: Choose a subset of teachers to judge the projects. There are 10 teachers, and you need to select 3 to be judges. The order of selection does not matter.

 The solution to this problem is provided below but you may want to first try yourself.

Solution:

1. Booth Assignment (Multiplication Rule):

For each project, you can choose any of the 5 booths. Since each project gets its own booth, we have:

• Project 1: 5 choices
• Project 2: 4 remaining choices (after assigning Project 1)
• Project 3: 3 remaining choices (after assigning Projects 1 and 2)

Total number of ways to assign the booths (using the multiplication rule): 5×4×3 = 60

2. Volunteer Scheduling (Factorials):

There are 4 volunteers and 4 different roles. There are 4 volunteers available for the first role. You can choose any one of them, so you have 4 choices. After assigning the first role, 3 volunteers remain. You can choose any one of them for the second role, giving you 3 choices. For the next two roles you will have 2 and 1 choices respectively. Therefore the number of ways to assign these roles is: 4! = 4×3×2×1 = 24

3. Selecting Judges (Combinations):

Out of 10 teachers, you need to select 3 to be judges. The order in which they are selected does not matter. The number of ways to choose the judges is: C (10, 3) = 120

This scenario demonstrates how different counting principles can be applied to various aspects of a single problem.

Intuition behind using Combination:

You might be rightfully wondering why we did not just use multiplication rule for finding the number of ways for the selection of judges. Let's explain the logic by considering how many choices you have at each step and why we divide by certain numbers.

1. Choosing the First Judge: You have 10 teachers to start with. So, there are 10 possible choices for the first judge.
   
   
2. Choosing the Second Judge: After selecting the first judge, 9 teachers remain. So, there are 9 possible choices for the second judge.
   
   
3. Choosing the Third Judge: After selecting the second judge, 8 teachers remain. So, there are 8 possible choices for the third judge.

Total Number of Selections (Considering Order):

If we consider the order in which we select the judges (which means we are considering different orders as different selections), we would multiply the number of choices at each step. This gives us the number of ways to select 3 judges considering different orders: 10×9×8 = 720

Adjusting for Order (Why We Divide):

However, in this scenario, the order in which we select the judges does not matter. For example, selecting Teachers A, B, and C is the same as selecting B, A, and C or C, B, and A. Each group of 3 judges can be arranged in 3×2×1 = 6 different ways (since there are 3 judges, the first can be any of the 3, the second any of the remaining 2, and the last is fixed).

To account for this and avoid counting the same group multiple times, we divide the total number of ordered selections by the number of ways to arrange 3 judges: 10×9×8 / 3×2×1= 120. So, there are 120 unique ways to choose 3 judges from 10 teachers when the order does not matter.

This division to handle non-unique groups is provided by the combination operator.