Question

A student has four examinations to write, and there are 10 examinations periods available. How many ways are there to schedule the examinations?

Short answer

There are 210 ways to schedule the examinations.

Step-by-step solution

Understanding the Problem

The problem asks us to determine in how many ways a student can schedule four examinations over ten different periods. This is essentially a problem of selecting positions for 4 exams out of 10 available periods.

Identifying the Formula

The problem is a combination problem where we want to find out in how many ways we can choose 4 periods from the 10 available. The formula for combinations is given by: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] where \( n \) is the total number of items to choose from, and \( k \) is the number of items to choose.

Applying the Formula

In this case, \( n = 10 \) and \( k = 4 \). Using the combination formula, we calculate:\[ \binom{10}{4} = \frac{10!}{4!(10-4)!} \]

Calculating Factorials

First, calculate the factorials involved:- \( 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \) (we'll simplify this calculation step by step)- \( 4! = 4 \times 3 \times 2 \times 1 = 24 \)- \( 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \).

Simplifying the Expression

To simplify \( \binom{10}{4} \):\[ \binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4!} = \frac{10 \times 9 \times 8 \times 7}{24} \]Calculate the numerator:\[ 10 \times 9 = 90 \]\[ 90 \times 8 = 720 \]\[ 720 \times 7 = 5040 \]Then divide by \( 24 \):\[ \frac{5040}{24} = 210 \].

Conclusion

Therefore, there are 210 different ways to schedule the four examinations within the 10 periods.

Key concepts

Combination Formula

In the world of combinatorics, the combination formula emerges as a fundamental tool. It is used when we need to determine how many ways we can choose a subset of items from a larger set without considering the order. In our example, we want to know how many ways we can select 4 exam periods out of 10 available slots. The formula used for combinations is:

• \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)

Here, \( n \) represents the total available items—in this case, examination periods—while \( k \) stands for the items to choose—in this instance, exams to schedule. Applying the combination formula gives you the total number of possible choices, simplifying the task without having to list all possible combinations. By placing emphasis on the choice rather than sequence, combinations overcome the complexity of order-sensitive calculations.

Factorials

Factorials are an essential component in the language of permutations and combinations, and are central to computing the combination formula. The factorial of a number, denoted by an exclamation mark (like \( n! \)), represents the product of all positive integers less than or equal to that number. For combinatorial problems, knowing how to calculate factorials is crucial:

• \( 4! = 4 \times 3 \times 2 \times 1 = 24 \)
• \( 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \)
• \( 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \)

Factorials grow exceedingly fast, making computational simplifications necessary when dealing with larger numbers. Fortunately, the combination formula often results in many terms cancelling out, significantly easing the burden of calculation.

Scheduling Problems

Scheduling problems often present themselves through combinatorial challenges. In essence, these problems require you to distribute resources or tasks over a set period or space effectively. In the given scenario, we have to place 4 exams into 10 available periods, illustrating a scheduling issue.

• First, identify how many tasks (exams) there are.
• Determine the available slots (periods) they can fit into.
• Apply a combinatorial approach to ascertain the number of viable schedules.

By understanding the number of ways to choose periods using combinations, students can visualize and devise effective scheduling solutions within certain parameters. This method, appropriate for scheduling exams, can also be applied to other real-life scenarios where time or resources need to be allocated efficiently, making combinatorics a versatile and practical field of mathematics.