An office manager needs to assign four offices to 12 visitin

An office manager needs to assign four offices to 12 visiting professors during the summer months. There is an office on the north-side of the Math Sciences Building, as well as an office on the south, east and west-sides of the Math Sciences Building. Three professors will be assigned to each office. Of the 12 visiting professors: five are from Europe, three are from Canada, and four are from the Middle East. If the assignment of offices is done randomly, In how many different ways can the assignment of offices take place? What is the probability that the three Canadians are assigned to office on the northside of the Math Sciences Building? What is the probability that two Europeans will be assigned to the south-side office and three Europeans will be assigned to the west-side office? What is the probability that all Canadian professors will be assigned to the same office?

Solution

a)

There are 12!/[3!3!3!3!] = 369600 ways to do the task.

b)

The three Canadians are fixed, so the remaining have 9!/[3!3!3!] = 1680 ways to do it.

Thus,

P = 1680 / 369600 = 0.004545455 [answer]

c)

There are 5!/[2!3!] = 10 ways to arrange the 5 europeans.

Meanwhile, there are 7!/[3!3!1!] = 140 ways to arrange the others.

Thus, there are 140*10 = 1400 ways to do that task.

Thus,

P = 1400/369600 = 0.003787879 [answer]

d)

The three Canadians are fixed, so the remaining have 9!/[3!3!3!] = 1680 ways to do it.

However, the three Canadians have 4 rooms to choose from, so a total of 1680*4 = 6720 ways.

Thus,

P(canadians in same room) = 6720/369600 = 0.018181818 [answer]