Question

In how many ways can 5 people all have different birthdays? Assume that there are 365 days in a year.

Short answer

7,839,725,680 ways

Step-by-step solution

Understand the Problem

Determine the number of ways 5 people can have different birthdays out of 365 days in a year.

Set Up the First Person

The first person can have their birthday on any of the 365 days of the year. Therefore, the number of options for the first person is 365.

Set Up the Second Person

The second person must have their birthday on any day except the one chosen by the first person. So, the number of available days is 364.

Set Up the Third Person

The third person must have their birthday on any of the remaining days except the two chosen by the first and second persons. So, the number of available days is 363.

Set Up the Fourth Person

The fourth person must have their birthday on any of the remaining days except the three chosen by the first three persons. So, the number of available days is 362.

Set Up the Fifth Person

The fifth person must have their birthday on any of the remaining days except the four chosen by the first four persons. So, the number of available days is 361.

Calculate the Total Number of Ways

Multiply the number of choices for each person: \[ 365 \times 364 \times 363 \times 362 \times 361 \]

Simplify the Result

Calculate the product: \[ 365 \times 364 \times 363 \times 362 \times 361 = 7,839,725,680 \]

Key concepts

Permutations

Permutations are an essential concept in solving problems like the birthday problem. They refer to the different ways in which a set of objects can be arranged. When considering permutations, it’s crucial to remember that order matters. For example, the order in which people are assigned their birthdays is significant.
In the given exercise, we calculate the number of ways 5 people can have different birthdays. Since the order in which each person gets their birthday matters, we are dealing with permutations. The formula to find the number of permutations of 5 people out of 365 days is:
\[ P(365, 5) = 365 \times 364 \times 363 \times 362 \times 361 \]
This multiplication accounts for the fact that each subsequent person has one less day to pick from than the previous person. This setup of multiplying sequential numbers is a hallmark of permutations.

Probability

Probability deals with the likelihood of an event occurring. When calculating the probability related to permutations, the number of successful outcomes is divided by the total number of possible outcomes.
In the context of the birthday problem, if we were to calculate the probability that 5 people have different birthdays, we would first find the total number of possible outcomes, which is simply the number of days to the power of the number of people involved:

• Total possible outcomes: \( 365^5 \)
• Number of successful outcomes where all 5 people have unique birthdays: \( 365 \times 364 \times 363 \times 362 \times 361 \)

The probability \( P \) can then be calculated as: \[ P = \frac{365 \times 364 \times 363 \times 362 \times 361}{365^5} \]
Understanding how to break down and compute both the numerator and denominator in probability problems is key to mastering this concept.

Combinatorics

Combinatorics is the branch of mathematics that deals with counting, both as a means and an end in obtaining results. It includes the study of permutations, combinations, and more to solve problems.
The birthday problem utilizes basic combinatoric principles to arrive at the solution. Each step in the exercise aligns perfectly with the combinatorial method of counting the number of ways events can occur. Key principles include:

• Permutations: Arranging 5 people among 365 days
• Sequential selection: Each person’s birthday reducing the available choices by one

These methods help break down complex scenarios into smaller, manageable parts, facilitating easier calculation of probabilities, permutations, and combinations.
Understanding combinatorics is crucial for tackling a wide range of problems in probability and statistics, paving the path toward comprehending more complex mathematical concepts.