combinatorics/probability problem -- please help

Question

I downloaded a file with some sample combinatorics and prob problems, and trying to solve this one:

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?

The answers sheet says it's 8/9. I get 1/5. Here's how I calculate this:

number of possible arrangements:

sec1 sec2 sec3
4 0 0 => 3 arrangements (3!/2!)
3 1 0 => 6 arrangements (3!)
2 2 0 => 3 arrangements
2 1 1 => 3 arrangements, and the only condition satisfying at least 1 report/secretary

arrangements total: 3+6+3+3=15
arrangements where each gets at least one report: 3

Probability of last arrangement: 3/15=1/5

So where am I going wrong? Or is the known answer for this problem wrong?

OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/there-are-th ... 31926.html

Narenn · Accepted Answer

Probability = favorable outcomes / Total Outcomes.

Total Outcomes =  Number of ways in which Reports can be assigned to the secretaries.
There are 4 reports and 3 secretaries and each report can be assigned to any one of the 3 secretaries
So First Report can be assigned in 3 ways, simiarly 2nd, 3rd, and 4th Report also can be assigned in 3 ways each
So total number of ways = 3 X 3 X 3 X 3 = 3^4 = 81

Favorable Outcomes =  Each Secretary should get atleast one Report.
3 Secretaries can get 4 Reports in 4P3 ways. Now 1 Report left. That can be allotted to any one of 3 secretaries in 3 ways.
Hence Total permutations = 4P3 X 3------->(4!/ (4-3)!) X 3----------> 4! X 3------------>24 X 3 = 72

So the desired Probability is 72/81 = 8/9

Regards,

Narenn.

PS :- This concept will be thoroughly covered in my upcoming article on Permutations and Combination. You can find it in my signature - perhaps on tomorrow.

Dixon · Answer

Thank you. I kinda get the explanation, hope to find some more similar problems to get a bit more practice and really grasp these concepts.

KarishmaB · Answer

The reason you are not getting the correct answer is that you are assuming the letters are identical. You say: sec1 sec2 sec3 3 1 0 => 6 arrangements (3!) ( Here, you get 6 arrangements becoz sec1 could have 3 or 1 or 0 letters and so on but which 3 letters does sec1 have? The 4 depts will have 4 different letters) Instead, try to solve it like this: Total no of ways of distributing 4 letters among 3 secretaries = 3*3*3*3 = 81 (such that some secretaries may get no letters) To ensure that each secretary should get at least 1 letter, the distribution will be 2, 1, 1. Select 2 letters out of the 4 and select one secretary out of the three to give the 2 letters => 4C2 * 3C1 Now, you have 2 letters to give to 2 secretaries => 2! Number of favorable ways = (4*3/2) * 3 * 2 Required probability = 36/81 Try out my blog. I have discussed grouping identical/distinct items in various ways on it. https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/categor ... om/page/2/ Scroll down to the last post. It is the first post on Combinatorics. The 10-12 posts above it are all on probability/combinatorics

infotalk · Answer

I dont think the answers stated here are correct. Correct answer is 39/81.

Total number of outcomes = 81.
Probability that each gets atleast 1 = 1 - probability that each doesnt get atleast one report

=================================
Probability that each doesn't get atleast one report
=================================

Let reports be W,X,Y,Z and secretaries be A,B,C.

Now we have to find ways that atleast one of the secretary has 0 reports assigned to him.

We will do this for one secretary (find the number of outcomes that secretary has for 0 reports assigned) and then add up for all 3 secretaries.

==================================================
Number of outcomes for which "A" doesnt get assigned any report (W,X,Y,Z). 
==================================================

A = 0, B = 1, C = 3
A = 0, B = 3, C = 1
A = 0, B = 2, C = 2

So above combination is the only way A is assigned 0 reports. We are not done here. For each of the above cases, there is more than one outcome associated with it.

e.g. A=0, B=1, C=3

B=W, C=XYZ
B=X, C = WYZ
B=Y, C=WXZ
B=Z, C=WXY

4 OUTCOMES. Multiply this by 2 since we have two cases of (1,3). =  8

e.g. A=0, B=2, C=2

B=WX
B=WY
B=WZ
B=XY
B=XZ
B=YZ

So there are  6  outcomes for (2,2) case.

So total number of outcomes for where secretary A does not get any report assigned = 8 + 6 =  14

Number of outcomes for which secretaries A, B & C all don't get any reports assigned = 14 * 3 = 42

Probability that atleast one secretary is assigned 0 reports = 42 / 81 = 1/2.

WE ARE NOT DONE YET.

We have not considered the case where all 4 reports are assigned to a single secretary. There would be 3 such outcomes.

So the probability that atleast 1 report is assigned to each secretary is 39/81.

vitorpteixeira · Answer

VeritasPrepKarishma,

Could you please explain why all the way are 3X3X3X3?

Bunuel · Answer

Each of the four reports can be assigned to any of the three secretaries, so each report has 3 options, total = 3*3*3*3. This question is discussed here: https://gmatclub.com/forum/there-are-th ... 31926.html Please continue discussion in that thread.

combinatorics/probability problem -- please help

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combinatorics/probability problem -- please help

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615 to 715 in 15 Days (Ria's Strategies)

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