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. a. (type an integer.)

Explanation:

Step1: Calculate total number of permutations

The total number of people is 7. The number of ways they can arrive (total number of permutations of 7 people) is given by the permutation formula $n!$, where $n = 7$. So the total number of ways they can arrive is $7!=7\times6\times5\times4\times3\times2\times1 = 5040$.

Step2: Fix Ian first and Larry last for part b

If Ian arrives first and Larry arrives last, then we only need to arrange the remaining 5 people in the middle. The number of ways to arrange $n$ distinct objects is $n!$. Here $n = 5$, so the number of ways is $5!=5\times4\times3\times2\times1=120$.

Step3: Calculate probability for part c

The number of favorable outcomes (Ian first and Larry last) is 120 (from step 2), and the total number of outcomes is 5040 (from step 1). The probability $P$ is the number of favorable outcomes divided by the total number of outcomes. So $P=\frac{120}{5040}=\frac{1}{42}$.

Answer:

a. 5040
b. 120
c. $\frac{1}{42}$