Question

ian, sergio, simone, jim, kim, eduardo, dawn, and larry have all been invited to a dinner party. they arrive randomly and each person arrives at a different time. a. in how many ways can they arrive? b. in how many ways can ian arrive first and larry last? c. find the probability that ian will arrive first and larry last.

Explanation:

Step1: Calculate total number of arrival - orders

The number of ways \(n\) people can arrive is given by the permutation formula \(n!\). Here, there are \(8\) people (Ian, Sergio, Simone, Jim, Kim, Eduardo, Dawn, and Larry), so the total number of ways they can arrive is \(n = 8!\)

$$8!=8\times7\times6\times5\times4\times3\times2\times1 = 40320$$

Step2: Calculate number of ways for Ian to arrive first and Larry to arrive last

If Ian arrives first and Larry arrives last, then we only need to arrange the remaining \(6\) people in the middle. The number of ways to arrange \(m\) distinct objects is \(m!\). Here \(m = 6\), so the number of ways to arrange the 6 people in the middle is \(6!\)

$$6!=6\times5\times4\times3\times2\times1=720$$

Step3: Calculate number of ways for Ian to arrive first

If Ian arrives first, then we need to arrange the remaining \(7\) people. The number of ways to arrange \(7\) people is \(7!\)

$$7!=7\times6\times5\times4\times3\times2\times1 = 5040$$

Step4: Calculate probability

The probability \(P\) that Ian arrives first and Larry arrives last is the number of favorable outcomes (Ian first and Larry last) divided by the total number of outcomes. The number of favorable outcomes is \(6!\) and the total number of outcomes is \(8!\). So \(P=\frac{6!}{8!}=\frac{6!}{8\times7\times6!}=\frac{1}{56}\)

Answer:

a. \(8! = 40320\) ways for them to arrive.
b. \(6! = 720\) ways for Ian to arrive first and Larry to last.
c. The probability that Ian will arrive first and Larry last is \(\frac{1}{56}\)