Question

How many ways can an eight-person committee be chosen from a group of 10 new members and 15 old members if the committee is composed of: (a) Four members from each group (b) More new members than old members (c) At least two new members

Short answer

(a) 286,650 ways. (b) 138,555 ways. (c) 1,010,790 ways.

Step-by-step solution

Determine Possible Combinations for Part (a)

We need to select 4 members from 10 new members and 4 members from 15 old members. This is a combinations problem and uses the combination formula \( \binom{n}{r} \), which represents the number of ways to choose \( r \) elements from \( n \) elements without regard to order.For new members: \( \binom{10}{4} \).For old members: \( \binom{15}{4} \). Thus, the total combinations are \( \binom{10}{4} \times \binom{15}{4} \). Calculate each combination:\[ \binom{10}{4} = \frac{10!}{4!(10-4)!} = 210 \]\[ \binom{15}{4} = \frac{15!}{4!(15-4)!} = 1365 \]Multiply these results:\( 210 \times 1365 = 286,650 \).

Solve Part (b) by Counting Valid Combinations

We require more new members than old members. The possible valid distributions are:* 5 new members, 3 old members: \( \binom{10}{5} \times \binom{15}{3} \)* 6 new members, 2 old members: \( \binom{10}{6} \times \binom{15}{2} \)* 7 new members, 1 old member: \( \binom{10}{7} \times \binom{15}{1} \)* 8 new members, 0 old members: \( \binom{10}{8} \times \binom{15}{0} \)Calculate each combination separately:\[ \binom{10}{5} \times \binom{15}{3} = 252 \times 455 = 114,660 \]\[ \binom{10}{6} \times \binom{15}{2} = 210 \times 105 = 22,050 \]\[ \binom{10}{7} \times \binom{15}{1} = 120 \times 15 = 1,800 \]\[ \binom{10}{8} \times \binom{15}{0} = 45 \times 1 = 45 \]Add these results:114,660 + 22,050 + 1,800 + 45 = 138,555.

Solve Part (c) Using a Minimum Member Requirement

The requirement is at least 2 new members. Subtract from the total all combinations with fewer than 2 new members, which are:* 0 new members, 8 old members: \( \binom{10}{0} \times \binom{15}{8} \)* 1 new member, 7 old members: \( \binom{10}{1} \times \binom{15}{7} \)Calculate each separately:\[ \binom{10}{0} \times \binom{15}{8} = 1 \times 6,435 = 6,435 \]\[ \binom{10}{1} \times \binom{15}{7} = 10 \times 6,435 = 64,350 \]Compute the total combinations without any conditions:\[ \binom{25}{8} = \frac{25!}{8!(25-8)!} = 1,081,575 \]Subtract to find combinations with at least 2 new members:1,081,575 - 6,435 - 64,350 = 1,010,790.

Key concepts

Combination Formula

In combinatorics, the combination formula is a key concept used to determine the number of ways to select a group of items from a larger pool, without considering the order of selection. The formula is expressed as: \[\binom{n}{r} = \frac{n!}{r!(n-r)!}\]Here, \(n\) represents the total number of available options, and \(r\) represents the number of selections to be made. The exclamation mark denotes a factorial, which is the product of all positive integers up to that number, such as \(4! = 4 \times 3 \times 2 \times 1 = 24\).
This formula is particularly useful in committee selection problems, where choices have to be made without regard for arrangement, such as in this exercise where the number of members from different groups are to be selected for a committee.

New Members

In the context of the problem, new members refer to the more recently added individuals in the selection pool. When forming committees, especially in a structured body like an organization or club, incorporating new members may be strategic or mandatory for inclusive and diverse participation.
In this exercise, there are 10 new members to choose from, and each question considers different scenarios about how many should be selected. These options might include having more new members than old ones or ensuring a minimum number of them are included.
Understanding the combination formula allows us to calculate the different possible committee configurations, highlighting the mathematical approach involved in making such selections.

Old Members

Old members in this scenario refer to individuals who have been a part of the organization longer. In terms of committee selection, seasoned members often bring valuable experience and insight, which is why their selection might also be a focal point in the exercise.
Given that there are 15 old members, the combination of how they're chosen, either balanced with new members or not, becomes a central part of the computation.
The selection of old members using the formula follows the same principle as with new members. For example, calculating \(\binom{15}{4}\) to select four old members involves understanding their significance and ensures their representation is maintained during committee formation.

Committee Selection

Committee selection involves choosing a group that effectively represents the interests and skills of its larger organization. This process can be complex due to different requirements for representation, such as having a balance of new and old members or meeting a minimum participation threshold.
In this problem, we handle scenarios like selecting equal numbers from both groups, ensuring a certain type surpasses the other, or meeting specific minimum numbers. Each scenario serves to illustrate the broader applicability of combinatorics in decision-making processes.
These considerations are solved using the combination formula, allowing students to calculate the number of possible committee formations. Thus, these exercises emphasize understanding the principle behind fair and strategic selection in organizational contexts.