Problem 67 - Algebra and Trigonometry

Question

Solve the problem using the appropriate counting principle(s). Dance Committee A school dance committee is to consist of two freshmen, three sophomores, four juniors, and five seniors. If six freshmen, eight sophomores, twelve juniors, and ten seniors are eligible to be on the committee, in how many ways can the committee be chosen?

Step-by-Step Solution

Answer

There are 104,588,400 ways to choose the dance committee.

Step 1: Understand the Task

The task is to choose a specific number of students from each grade: 2 freshmen, 3 sophomores, 4 juniors, and 5 seniors, from the eligible students in each respective grade.

Step 2: Apply Combination Formula for Freshmen

To choose 2 freshmen from 6 eligible ones, use the combination formula: \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \). So here: \( \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6!}{2!4!} = 15 \).

Step 3: Apply Combination Formula for Sophomores

To choose 3 sophomores from 8 eligible ones, use the combination formula: \( \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!} = 56 \).

Step 4: Apply Combination Formula for Juniors

To choose 4 juniors from 12 eligible ones, use the combination formula: \( \binom{12}{4} = \frac{12!}{4!(12-4)!} = \frac{12!}{4!8!} = 495 \).

Step 5: Apply Combination Formula for Seniors

To choose 5 seniors from 10 eligible ones, use the combination formula: \( \binom{10}{5} = \frac{10!}{5!(10-5)!} = \frac{10!}{5!5!} = 252 \).

Step 6: Combine All Combinations

The total number of ways to choose the committee is the product of the individual combinations: \( 15 \times 56 \times 495 \times 252 \).

Step 7: Calculate the Final Answer

Now multiply these values: \( 15 \times 56 = 840 \), \( 840 \times 495 = 415800 \), and finally \( 415800 \times 252 = 104588400 \).

Step 8: Conclusion

Thus, the total number of ways to choose the dance committee is 104,588,400.

Key Concepts

Combination Formula

When solving problems in combinatorics that require choosing a specific number of items (or people, like in a committee) from a larger group, the combination formula is our go-to tool. It helps calculate how many possible ways there are to choose "r" items from a larger set of "n" items, and is written as: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]The "!" symbol stands for factorial, which we'll explore in more detail later.

• This formula is key because it considers how choosing items is different from arranging them; combinations don't care about order, only selection.
• For example, in the problem provided, choosing 2 freshmen from 6 is a combination problem since the order in which you choose them doesn’t matter.

Every category of students (freshmen, sophomores, juniors, seniors) involves a similar calculation using this combination formula.

Counting Principles

Count principles in mathematics help us determine the total number of outcomes for various events. By applying them, we find how choices for different groups combine into a larger set of options.

• Usually, we use the multiplication principle of counting, which says if there are "a" ways to do something and "b" ways to do another thing, then there are \( a \times b \) ways to do both.
• In our committee problem, the separate counts of ways to choose each class are multiplied because each selection is independent - the number of ways to choose freshmen does not affect how we choose sophomores, juniors, or seniors.

Thus, the total number involves multiplying the combination outcomes for each year: \[ 15 \times 56 \times 495 \times 252 \]This shows the cumulative possibilities across all groups.

Factorial Calculation

Factorials, symbolized as an exclamation mark "!", are crucial in calculating combinations. A factorial represents the product of all positive integers up to a specified number. For instance, \( n! \) means \( n \times (n-1) \times (n-2) \times \,\ldots\, \times 1 \).Let's explore how factorials are used in the combination formula and, hence, the problem:

• To calculate \( \binom{6}{2} \), for example, we need \( 6! \), \( 2! \), and \( 4! \). These are factorials of 6, 2, and 4. Each breaks down into its constituent multiplications.
• Moreover, factorials help in understanding why order doesn’t matter in combinations since \( r! \) and \( (n-r)! \) divide out the order-associated variations.

Understanding how to compute and apply factorials helps ensure accuracy when working with combinations within counting problems.