Problem 35 - Basic Statistics for Business and Economics with Student

Question

There are 10 members on the Community Appearance Board of Los Angeles County. This board is charged with ensuring that all new construction projects meet county appearance standards. A new terminal is to be added to the Los Angeles International Airport. A subcommittee of four members is to be created to review the initial drawing of the project. How many different subcommittees are possible?

Step-by-Step Solution

Answer

There are 210 different possible subcommittees.

Step 1: Understanding the problem

We need to form a subcommittee of 4 members from a total of 10 members. The order in which we select the members does not matter, only the combination. Thus, this is a combinations problem.

Step 2: Applying the combinations formula

To find how many combinations of 4 members can be made from 10, we use the combinations formula: \[ C(n, r) = \frac{n!}{r!(n-r)!} \]where \(n\) is the total number of items to choose from, and \(r\) is the number of items to choose.

Step 3: Substituting the values

Here, \(n = 10\) (the total number of board members) and \(r = 4\) (the number of members in each subcommittee). Substituting these into the formula gives:\[ C(10, 4) = \frac{10!}{4!(10-4)!} \].

Step 4: Calculating factorial values

Calculate the factorials:- \(10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\) - \(4! = 4 \times 3 \times 2 \times 1\)- \(6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1\)

Step 5: Simplifying the expression

Now, substitute these factorial values into the formula:\[ C(10, 4) = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} \]Cancel out the common factors and compute the result.

Step 6: Final Calculation

Calculating the expression gives:\[ C(10, 4) = \frac{5040}{24} = 210 \]So, there are 210 different possible subcommittees.

Key Concepts

Combinations

Combinations are a way to choose a subset from a larger set without regard to the order of selection. Imagine you are making a four-person subcommittee from ten people. It doesn't matter who's picked first, second, third, or fourth; only the final group of four matters. To find the number of combinations, we use the formula:

• \( C(n, r) = \frac{n!}{r!(n-r)!} \)
• Here, \(n\) is the total number, and \(r\) is the number of selections.

By calculating combinations, we can determine how many unique groups can be formed from a larger set. This concept is very useful in scenarios where the sequence of selection doesn't matter, such as choosing committee members.

Factorial Calculation

The calculation of a factorial, denoted by an exclamation point (!), means multiplying a series of descending natural numbers. For example, \(10!\) means 10 times 9 times 8 and so on down to 1. It's like a countdown multiplication:

• \(10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\)
• \(4! = 4 \times 3 \times 2 \times 1 \)
• \(6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 \)

These calculations can seem complex at first, but they simplify the formula for combinations and permutations. Luckily, you don't always have to multiply these out by hand—calculators or software can handle these efficiently.

Subcommittee Selection

The exercise in question is a perfect illustration of subcommittee selection using combinations. When forming a subcommittee, you're looking to select a group of individuals from a larger pool. Here, the goal was to select 4 members from a 10-member board. To find out how many unique groups of 4 can be formed, you apply the combinations formula.For this particular problem, the combinations formula tells us there are 210 different ways to select the subcommittee of four from the ten:

• Substitute \(n = 10\) and \(r = 4\) into the combinations formula \( C(10, 4) = \frac{10!}{4!(10-4)!} \)
• The result is 210 unique subcommittees.

This concept is pivotal in many decision-making processes, ensuring each possible selection is considered without duplication of effort.