Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

combinatorics/probability problem -- please help [#permalink]

Show Tags

03 Apr 2013, 17:53

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?

Re: combinatorics/probability problem -- please help [#permalink]

Show Tags

07 Apr 2013, 11:21

3

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

Dixon wrote:

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?

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. _________________

Re: combinatorics/probability problem -- please help [#permalink]

Show Tags

09 Apr 2013, 23:12

Expert's post

Dixon wrote:

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?

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

Re: combinatorics/probability problem -- please help [#permalink]

Show Tags

25 Mar 2014, 00:00

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.

Re: combinatorics/probability problem -- please help [#permalink]

Show Tags

18 Jul 2015, 11:30

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

http://blog.ryandumlao.com/wp-content/uploads/2016/05/IMG_20130807_232118.jpg The GMAT is the biggest point of worry for most aspiring applicants, and with good reason. It’s another standardized test when most of us...

I recently returned from attending the London Business School Admits Weekend held last week. Let me just say upfront - for those who are planning to apply for the...