Problem 46 - College Algebra with Corequisite Support

Question

How many ways can a committee of 3 freshmen and 4 juniors be formed from a group of 8 freshmen and 11 juniors?

Step-by-Step Solution

Answer

There are 18,480 ways to form the committee.

Step 1: Understand the Problem

We need to form a committee consisting of 3 freshmen and 4 juniors from a group of 8 freshmen and 11 juniors. This is a combinatorics problem where we have to find the number of combinations.

Step 2: Calculate Combinations of Freshmen

To select 3 freshmen from a group of 8 freshmen, we use the combination formula: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \] \Applying it here: \[ \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \] \So, there are 56 ways to choose 3 freshmen.

Step 3: Calculate Combinations of Juniors

To select 4 juniors from a group of 11 juniors, again use the combination formula: \[ \binom{11}{4} = \frac{11!}{4!(11-4)!} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330 \] \Therefore, there are 330 ways to choose 4 juniors.

Step 4: Calculate Total Combinations

The total number of ways to form the committee is the product of the ways to choose freshmen and juniors, as these choices are independent: \[ 56 \times 330 = 18480 \] Thus, there are 18,480 ways to form the committee.

Key Concepts

Combinations

Combinations are a fundamental concept in combinatorics where the order of selection does not matter. When choosing a subset of items from a larger set, combinations help us calculate the number of possible selections without regard to the order in which items are chosen. For example, choosing 3 students out of 8 does not consider whether you pick Anna first or Mark; it only matters who is chosen. This is why, when forming a committee, we use combinations instead of permutations, which would consider the order significant.

When calculating combinations, we often use the binomial coefficient, which involves factorials to determine the number of ways to choose a subset from a set. Understanding which real-world problems require combination calculations as opposed to permutation solves half the problem in combinatorics.

Factorial

The concept of factorial is essential in calculating combinations. Denoted by an exclamation mark (such as "n!"), it represents the product of all positive integers up to "n." For instance, if you have n = 5, the factorial is 5! = 5 × 4 × 3 × 2 × 1 = 120. Factorials are crucial because they account for all possible arrangements of a set, which is a basic necessity when determining combinations.

The factorial provides a method to calculate large numbers quickly, allowing us to use these in the binomial coefficient formula. Factorials become particularly important as the numbers grow, hence the need for a structured approach when counting possibilities in larger sets.

Committee Formation

Committee formation problems like the one in this exercise require picking a precise number of members from different groups. Here, we're selecting 3 freshmen from 8 and 4 juniors from 11. Such problems are common in organizing, planning, and managing tasks where specific roles or slots must be filled.

These problems often highlight the real-world application of mathematical concepts and demonstrate how understanding combinations and factorials can directly impact practical scenarios. When assembling a diverse committee, you compute separately for each subgroup and then multiply to find the total combinations possible. This multiplication reflects the fact that selecting different subgroups are independent events.

Binomial Coefficient

The binomial coefficient is a numerical symbol often written as \( \binom{n}{r} \), used to calculate the number of ways to choose "r" items from a total of "n" without considering order. It's computed by the formula \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \). This coefficient simplifies combination calculations by using factorials of the total items and the chosen items.

The binomial coefficient is tremendously important because it helps simplify computations for large sets quickly. In our problem, it calculates both the freshmen and juniors in separate instances. Once you understand how to apply this coefficient, you can solve a wide range of problems involving combinations, showing its versatility and usefulness in the world of combinatorics.