Problem Solving

There are three secretaries who work for four departments. If each of the four departments has one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretaries are assigned at least one report?

OA -- [spoiler]4/9[/spoiler]

I'm fully aware of how to solve it by the regular method. Total number of possibilities = 3^4 = 81 and total number of desired outcomes = 6*6, giving us the final answer of [spoiler]4/9[/spoiler]

I'm looking for help in solving this one using the alternate method, which is:

1 - (probability not all 3 secretaries are assigned one report in 4 departments)

I know we have to calculate the following outcomes:

(A) Either one of the three typists is assigned reports for all 4 departments (That will be 3 choices)
(B) One of the typists is assigned a report for 2 departments and another typist is assigned reports for the other 2 departments, while the third typist gets nothing. (3 choices)
(C) One typist gets reports assigned for three departments and another gets assigned reports for 1 department, and the third gets nothing. (3 choices again)

And that's as far as I could go, before my brain stopped functioning.

I hope I can get some expert help in continuing where I left off on the ALTERNATE METHOD. Many thanks in advance.

knight247 wrote:There are three secretaries who work for four departments. If each of the four departments has one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretaries are assigned at least one report?

Good cases = all possible cases - bad cases.

All possible cases:
Number of options for the 1st report = 3. (Any of the 3 secretaries.)
Number of options for the 2nd report = 3. (Any of the 3 secretaries.)
Number of options for the 3rd report = 3. (Any of the 3 secretaries.)
Number of options for the 4th report = 3. (Any of the 3 secretaries.)
To combine these options, we multiply:
3*3*3*3 = 81.

A bad case occurs when all 3 secretaries DO NOT each receive at least 1 report.

Bad Case 1: 2 secretaries each receive a pair of reports
From the 3 secretaries, the number of ways to choose 2 = 3C2 = (3*2)/(2*1) = 3.
From the 4 reports, the number of ways to choose 2 for the 1st secretary = 4C2 = (4*3)/(2*1) = 6.
The remaining 2 reports must be assigned to the other selected secretary.
To combine these options, we multiply:
3*6 = 18.

Bad Case 2: 1 secretary receives 3 reports, another receives 1 report
From the 3 secretaries, the number of ways to choose 2 = 3C2 = (3*2)/(2*1) = 3.
From the 2 selected secretaries, the number of ways to choose 1 to receive 3 reports = 2. (Either of the 2 selected secretaries.)
From the 4 reports, the number of ways to choose 3 for this secretary = 4C3 = (4*3*2)/(3*2*1) = 4.
The 1 remaining report must be assigned to the other selected secretary.
To combine these options, we multiply:
3*2*4 = 24.

Bad Case 3: 1 secretary receives all 4 reports
Number of options for this secretary = 3. (Any of the 3 secretaries.)

Thus:
Good cases = 81-18-24-3 = 36.

Thus:
P(each secretary receives at least 1 report) = (good cases)/(all possible cases) = 36/81 = 4/9.

ALTERNATE... Perhaps a shorter one

Since 2 Reports will have to go to one Secretary so lets make the group of 2 books and consider the group as one entity only which goes to one secretary. The group of 2 books can be made in 4C2 ways = 6 ways
NOW
The First report can be assigned in 3 ways
Second report can be assigned in 2 ways
Third report can be assigned in 1 way and the last report can be assigned in 3 ways
So total ways to assign 4 reports to three secretaries such that each gets atleast one report = 3x2x1x4C2 = 36

Probability = 36/81 = 4/9