Question

How many ways can a committee of three men and two women be chosen from six men and four women? What if Adam Smith and Abigail Smith will not serve on the same committee?

Short answer

There are 105 ways to choose the committee under the given conditions.

Step-by-step solution

Choose Men

First, we need to choose 3 men from the available 6. This is a combination problem where the order does not matter. We can calculate this using the combination formula \( \binom{n}{r} \) where \( n \) is the total number and \( r \) is the number to choose. Here, \( \binom{6}{3} \) calculates the combinations.\[ \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]

Choose Women

Next, we choose 2 women from 4 available women. Again, we will use the combination formula \( \binom{n}{r} \). Thus, \( \binom{4}{2} \) needs to be calculated.\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \]

Total Combinations Without Restriction

Now, multiply the number of ways to choose men and women to find the total number of unrestricted combinations for the committee.\[ 20 \times 6 = 120 \]

Consider Restriction with Adam and Abigail

If Adam Smith and Abigail Smith will not serve together, first calculate the number of committees where they serve together. Consider Adam and Abigail are already chosen, then choose 1 man from the remaining 5 men, and 1 woman from the remaining 3 women.For 1 man from 5: \( \binom{5}{1} = 5 \)For 1 woman from 3: \( \binom{3}{1} = 3 \)Multiply the two results for the restricted part:\[ 5 \times 3 = 15 \]

Apply Restriction to Total

Subtract the number of restricted committees from the total unrestricted committees to get the final count of acceptable committees.\[ 120 - 15 = 105 \]

Key concepts

Combination formula

In combinatorics, the combination formula is a powerful tool that helps us determine the number of ways we can choose a subset of items from a larger set. This is often needed when the order of selection does not matter. The formula for combinations is given by:\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]where:

• \( n \) is the total number of items.
• \( r \) is the number of items to choose.
• \( n! \) denotes the factorial of \( n \), which is the product of all positive integers up to \( n \).

For instance, when choosing 3 men from 6 available men, we have \( n = 6 \) and \( r = 3 \), resulting in:\[ \binom{6}{3} = \frac{6!}{3!(6-3)!} = 20 \]The formula simplifies selection processes in scenarios such as picking individuals for committees, as we will see below.

Committee selection

When forming a committee from a group of people, the order in which members are chosen is usually not important. This turns the selection process into a combinatorial problem, perfectly suited for the combination formula. In the given exercise, the task is to form a committee consisting of 3 men and 2 women from a total of 6 men and 4 women. Let's break it down:

• Choosing 3 men from 6 men comes down to calculating \( \binom{6}{3} \), giving us 20 ways.
• Similarly, choosing 2 women from 4 women requires calculating \( \binom{4}{2} \), resulting in 6 ways.

To determine the total number of ways to form the committee without any restrictions, multiply the two results: \[ 20 \times 6 = 120 \]This shows how combinatorial logic simplifies the task of determining possible configurations for group selections.

Combinatorial restrictions

In many real-world applications of combinatorics, certain restrictions apply, complicating the selection process. In the exercise, Adam Smith and Abigail Smith cannot serve on the same committee. This introduces a restriction we need to adjust for in our total combination count.To handle this, we first calculate the number of unacceptable committees where Adam and Abigail serve together:

• Assuming Adam and Abigail are already chosen, select 1 additional man from the remaining 5 men, yielding \( \binom{5}{1} = 5 \) options.
• For the women, select 1 additional woman from the remaining 3, resulting in \( \binom{3}{1} = 3 \) choices.
• Combine these to find the total number of committees where Adam and Abigail serve together: \[ 5 \times 3 = 15 \] unacceptable committees.

To find the number of committees that respect the restriction, subtract these 15 from the total number of possible committees \( 120 \), giving:\[ 120 - 15 = 105 \]This adjusted total ensures all combinations meet the criteria, highlighting the importance of carefully considering combinatorial restrictions in problem-solving.