Question

Amy, Jean, Keith, Tom, Susan, and Dave have all been invited to a birthday party. They arrive randomly and each person arrives at a different time. In how many ways can they arrive?

In how many ways can Jean arrive first and Keith last? What is the probability that Jean will arrive first and Keith will arrive last?

Answer 1

Completely random = 720 ways, First and Last set = 24 ways, Probability of First and Last happening `=1/30`

Explanation:

We have several people arriving to a birthday party in a random order but all at different times. In how many ways can they arrive?

This is a Permutation question - we care about the order in which they arrive. The general equation for Permutations is:

`P_(n,k)=(n!)/((n-k)!)`, where n is the population we can pick from and k is the number of choices we're going to make.

We have 6 people to choose from and we're also selecting all 6:

`P_(6,6)=(6!)/((6-6)!)=(6!)/(0!) =(6!)/1=720` ways

How many ways can we have Jean arrive first and Keith last? That means we can have the other four show up at any time, and so we can calculate it this way:

`P_(4,4)=(4!)/((4-4)!)=(4!)/(0!) =(4!)/1=24` ways

The probability of Jean showing up first and Keith last is the ratio of the number of ways it can happen (24) divided by the number of ways they can all show up (720):

`24/720=1/30`