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Alternate method required

by knight247 » Wed Nov 26, 2014 10:51 am
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. :-D

I hope I can get some expert help in continuing where I left off on the ALTERNATE METHOD. Many thanks in advance.
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by GMATGuruNY » Wed Nov 26, 2014 11:46 am
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.
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by GMATinsight » Thu Nov 27, 2014 6:36 am
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
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