Question

tyrone, kim, maria, katrina, jim, eduardo, simone, and ian 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 tyrone arrive first and ian last?
c. find the probability that tyrone will arrive first and ian last.
a. (type an integer)

Explanation:

Step1: Calculate the number of ways for part a

The number of permutations of \(n\) distinct objects is \(n!\). Here \(n = 8\) (Tyrone, Kim, Maria, Katrina, Jim, Eduardo, Simone, and Ian).
The formula for permutations of \(n\) objects is \(P(n,n)=\frac{n!}{(n - n)!}=n!\)
So, \(n!=8! = 8\times7\times6\times5\times4\times3\times2\times1=40320\)

Step2: Calculate the number of ways for part b

If Tyrone arrives first and Ian last, then we only need to arrange the remaining \(n-2=6\) people.
The number of permutations of \(6\) distinct objects is \(6!\)
\(6! = 6\times5\times4\times3\times2\times1=720\)

Step3: Calculate the probability for part c

The probability \(P\) of an event is given by \(P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\)
The number of favorable outcomes (Tyrone first and Ian last) is \(6!\), and the total number of outcomes is \(8!\)
\(P=\frac{6!}{8!}=\frac{6!}{8\times7\times6!}=\frac{1}{8\times7}=\frac{1}{56}\)

Answer:

a. \(40320\)
b. \(720\)
c. \(\frac{1}{56}\)