Problem 42 - Algebra and Trigonometry

Question

In how many ways can a committee of 3 professors be formed from a department that has 8 professors?

Step-by-Step Solution

Answer

56

Step 1: Understand the Problem

The task is to determine the number of ways to select a committee of 3 professors from a department with 8 professors. This is a combination problem where the order of selection does not matter.

Step 2: Apply the Combination Formula

To solve the combination problem, use the combination formula: \[ C(n, k) = \frac{n!}{k! \, (n - k)!} \]where \( n \) is the total number of professors (8) and \( k \) is the number of professors to be chosen (3).

Step 3: Calculate the Factorials

Calculate the factorials involved in the formula:\[ n = 8 \implies 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320 \]\[ k = 3 \implies 3! = 3 \times 2 \times 1 = 6 \]\[ n - k = 5 \implies 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]

Step 4: Substitute and Simplify

Substitute the factorial values into the combination formula and simplify:\[ C(8, 3) = \frac{8!}{3! \, 5!} = \frac{40320}{6 \, \times \, 120} = \frac{40320}{720} = 56 \]

Step 5: Conclusion

The number of different ways to form a committee of 3 professors from 8 professors is 56.

Key Concepts

factorials

Factorials are mathematical expressions represented by an exclamation mark (!). The factorial of a positive integer is the product of all positive integers less than or equal to that integer. For example, 5 factorial (written as 5!) is calculated as:
5! = 5 × 4 × 3 × 2 × 1 = 120
Factorials are fundamental in permutations and combinations because they allow us to calculate the number of ways to arrange or select items from a set. Understanding how to compute and apply factorials is essential for solving these types of problems.

committee selection

Committee selection is a common problem in combinatorics. When forming a committee from a larger group, the order in which members are selected doesn't matter. This means we use combinations rather than permutations. Consider a department with 8 professors and the task of forming a committee of 3. The key is to determine the number of different groups (committees) that can be formed without regard to the order in which they are chosen.
To solve this, we use the combination formula, which accounts for the non-importance of order.

combination formula

The combination formula is used to find the number of ways to choose a subset of items from a larger set, without considering the order. The formula is:
\[ C(n,k) = \frac{n!}{k!(n-k)!} \]
Where:

• n is the total number of items (in this case, 8 professors).
• k is the number of items to be chosen (in this case, 3 professors).
• n!, k!, and (n-k)! are the factorials of n, k, and (n-k) respectively.

Substituting the values into this formula, we get:\[ C(8,3) = \frac{8!}{3!(8-3)!} = \frac{40320}{6 \times 120} = 56 \]This calculation tells us there are 56 different ways to form a committee of 3 professors from 8.

permutations and combinations

In mathematics, permutations and combinations are two ways of counting arrangements and selections of items. The key difference is whether the order matters.

• Permutations: These are arrangements where the order is important. For example, the arrangement ABC is different from BAC.
• Combinations: These are selections where the order does not matter. For example, selecting {A, B, C} is the same as selecting {C, B, A}.

When dealing with problems like committee selection, where the order of selection does not matter, we use combinations. Understanding these concepts helps in accurately determining the number of ways to arrange or select items in different scenarios.