Question

You invite five friends for dinner but forget to ask for a reply. a) What are the possible cases for the number of dinner guests? b) How many combinations of your friends could come for dinner? c) How does your answer in part b) relate to Pascal's triangle?

Short answer

a) 0 to 5 guests. b) 32 combinations. c) Sum of 5th row in Pascal's triangle equals 32.

Step-by-step solution

Understand the problem

Five friends are invited to dinner, but it's unknown how many will come. The problem has three parts: determining possible cases for the number of guests, counting combinations of friends that might come, and relating the result to Pascal's triangle.

- Determine possible cases for the number of guests

Since each of the 5 friends can either come or not come, the possible number of dinner guests includes any number from 0 to 5.

- Use combinations to count the possible groups of friends

Each possible group of friends that could come corresponds to selecting a subset of the 5 friends. This can be represented by combinations, where the number of ways to select k friends out of 5 is given by the binomial coefficient \(\binom{5}{k}\).

- Calculate the total number of combinations

The total number of possible combinations of friends attending is \(\sum_{k=0}^{5} \binom{5}{k} \). Using the binomial theorem, the sum of these coefficients is \(2^5 = 32\). Therefore, there are 32 possible combinations of friends who might come for dinner.

- Relate the answer to Pascal's triangle

Pascal's triangle displays binomial coefficients. The total number of combinations, 32, corresponds to the sum of the entries in the 5th row of Pascal's triangle (start counting from row 0), which is \(\sum_{k=0}^{5} \binom{5}{k} = 2^5 = 32\).

Key concepts

Binomial Coefficient

In mathematics, the binomial coefficient \binom{n}{k} represents the number of ways to choose k items from n items without regard to order. It's a fundamental concept used in combinatorics and probability. The value of \binom{n}{k} is calculated using the formula: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] where ! denotes factorial, meaning the product of all positive integers up to that number.
For example, with 5 friends and selecting any number of them to attend the dinner, the binomial coefficient \binom{5}{k} gives the number of possible groups of k friends that could come.
Therefore, for k = 2 friends, we have \[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = 10 \] showing there are 10 different combinations of 2 friends from the group of 5.

Pascal's Triangle

Pascal's triangle is a triangular array of the binomial coefficients. Each row represents the coefficients in the expansion of a binomial raised to a power.
To build Pascal's triangle, start with 1 at the top, then continue placing numbers below it in a pyramidal structure. Every number is the sum of the two numbers directly above it from the previous row.
For example, the first few rows are:

• Row 0: 1
• Row 1: 1, 1
• Row 2: 1, 2, 1
• Row 3: 1, 3, 3, 1
• Row 4: 1, 4, 6, 4, 1
• Row 5: 1, 5, 10, 10, 5, 1

The 5th row corresponds to the problem, where the values represent the number of ways to choose 0 to 5 friends: \[ \binom{5}{0}, \binom{5}{1}, \binom{5}{2}, \binom{5}{3}, \binom{5}{4}, \binom{5}{5} \] Summing these gives the total number of combinations: 32.

Binomial Theorem

The binomial theorem gives the expanded form of powers of binomials. It states that for any positive integer n and any numbers a and b, \[ (a+b)^n = \binom{n}{0}a^n b^0 + \binom{n}{1}a^{n-1}b^1 + \binom{n}{2}a^{n-2}b^2 + \text{...} + \binom{n}{n}a^0 b^n \] This theorem helps in understanding how combinations are distributed in a binomial expansion.
Relating to the problem, for n = 5 and a = b = 1, the formula becomes: \[ (1+1)^5 = \binom{5}{0}1^5 + \binom{5}{1}1^4 + \binom{5}{2}1^3 + \binom{5}{3}1^2 + \binom{5}{4}1^1 + \binom{5}{5}1^0 \] Simplifying, we get: \[ 2^5 = 32 \] This confirms the sum of the binomial coefficients, which is the total number of combinations of friends who might come for dinner.