Dawn comma Sergio comma Tyrone comma and Jim 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 Dawn arrive first and Jim last? c. Find the probability that Dawn will arrive first and Jim last.

a. 24

b. 2

c. 0.0833 = 8.33%

Step-by-step explanation:

a.

The first "slot" of person to arrive has 4 possibilities, then the second "slot" will have 3 possibilities, as one has already arrived, then the third "slot" has 2 possibilities, and the fourth "slot" has just 1 possibility.

So, multiplying all these combinations, we have 4*3*2*1 = 24 possible ways they can arrive

b.

If the first and the last person are already "locked", we just have possibilities for the second and third person. The second will have 2 possibilities (Sergio or Tyrone), and the third will have only 1 (the person that wasn't the second between Sergio and Tyrone). So, the number of possibilities is 2*1 = 2

c.

If we have 2 cases where Dawn is first and Jim is last, from a total of 24 possible cases, the probability is 2/24 = 1/12 = 0.0833 = 8.33%