Committee Formation Explained

This question requires us to determine the total number of unique or distinct committees that can be formed. We are given a total group of 9 people, and we need to select 3 members to form a committee. When forming a committee, the order in which the members are selected does not matter. For instance, if we select person A, then B, then C, it forms the same committee as selecting person C, then B, then A. Because the order of selection is not important, this is a classic problem of combinations.

Combinations: The Mathematical Concept

A combination refers to the selection of items from a larger set where the order of selection is not considered. The formula used to calculate the number of combinations of choosing \( k \) items from a set of \( n \) distinct items is given by:

\( C(n, k) = \binom{n}{k} = \frac{n!}{k!(n-k)!} \)

Let's break down the components of this formula:

• \( \mathbf{n} \): Represents the total number of available items in the larger group (in our case, the total number of people).
• \( \mathbf{k} \): Represents the number of items we need to choose for our smaller group or committee (in our case, the number of members in the committee).
• \( \mathbf{!} \): Denotes the factorial operation. For any positive integer \( x \), \( x! \) is the product of all positive integers less than or equal to \( x \). For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).

Calculating Distinct Committees Formed

In this specific problem, we have the following values:

• Total number of people (\( n \)) = 9
• Number of members to be selected for the committee (\( k \)) = 3

Now, we will substitute these values into the combination formula:

\( C(9, 3) = \frac{9!}{3!(9-3)!} \)

\( C(9, 3) = \frac{9!}{3!6!} \)

To solve this, we can expand the factorials. Remember that \( 9! = 9 \times 8 \times 7 \times 6! \). This allows us to simplify the expression by canceling out \( 6! \) from the numerator and the denominator:

\( C(9, 3) = \frac{9 \times 8 \times 7 \times 6!}{ (3 \times 2 \times 1) \times 6! } \)

\( C(9, 3) = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} \)

Next, we perform the multiplication in the numerator and the denominator:

Numerator: \( 9 \times 8 \times 7 = 72 \times 7 = 504 \)

Denominator: \( 3 \times 2 \times 1 = 6 \)

Finally, we divide the numerator by the denominator:

\( C(9, 3) = \frac{504}{6} \)

\( C(9, 3) = 84 \)

Therefore, there are 84 distinct committees that can be formed from a group of 9 people, choosing 3 members.

Summary for Committee Selection

Here is a quick overview of the process used to solve this type of committee selection problem:

The final calculation shows that 84 distinct committees can be formed.