combinatorics/probability problem -- please help

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

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

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Probability that each doesn't get atleast one report
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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.

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

VeritasPrepKarishma,

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

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.