Question

simone, sarah, sergio, maria, kim, and tyrone 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 simone arrive first and tyrone last?
c. find the probability that simone will arrive first and tyrone last.

a. (type an integer)

Explanation:

Step1: Calculate the number of ways they can arrive (a)

The number of permutations of \(n\) distinct objects is \(n!\). Here \(n = 6\) (Simone, Sarah, Sergio, Maria, Kim, and Tyrone).

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

Step2: Calculate the number of ways Simone arrives first and Tyrone last (b)

If Simone is first and Tyrone is last, then we need to arrange the remaining \(6 - 2=4\) people. The number of permutations of \(n\) distinct objects is \(n!\). Here \(n = 4\)

$$4! = 4\times3\times2\times1 = 24$$

Step3: Calculate the probability (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 is \(24\) (from part b) and the total number of outcomes is \(720\) (from part a)

$$P=\frac{24}{720}=\frac{1}{30}$$

Answer:

a. \(720\)
b. \(24\)
c. \(\frac{1}{30}\)