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# There are three secretaries who work for four departments.

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There are three secretaries who work for four departments.  [#permalink]

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Updated on: 19 May 2014, 06:45
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95% (hard)

Question Stats:

37% (02:54) correct 63% (02:50) wrong based on 484 sessions

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

A. 8/9
B. 64/81
C. 4/9
D. 16/81
E. 5/9

Originally posted by mahesh004 on 15 Jul 2006, 00:02.
Last edited by Bunuel on 19 May 2014, 06:45, edited 1 time in total.
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Re: There are three secretaries who work for four departments.  [#permalink]

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10 Feb 2015, 20:23
3
9
Lapetiteflo wrote:
Hi,
I have Big troubles with this question.
We havé total possibilities N =3^4
Then we can say that the probability to give at least one report to each secretary is the opposite of giving no report to at least one secretary.
So evaluating this proba :
-ways to choose 1 secretary among 3 (she will not be assigned any report)=3
-ways to affect 4 reports to 2 secretaries (it can be just one if ever) : examples (1122) assigns 2 first reports to secretary 1 and two last one to secretary 2. There are 2^4 possibilities (1111,1112,1122,....)

So result is 1-(3*2^4)/3^4=11/27.

The true result is 12/26.... Could you explain where is my mistake ?
Thanks a lot
Marie

"We havé total possibilities N =3^4"

What are these total possibilities? They are the ways in which the 4 reports can be distributed among 3 secretaries so that all reports may go to one secretary, the reports may be distributed among 2 secretaries or they may be distributed among all 3. So such cases are included (4, 0, 0), (0, 4, 0), (2, 2, 0), (1, 3, 0), (1, 1, 2), etc

In how many ways can you give 4 reports to only 1 secretary? You choose the secretary who will get the reports in 3C1 = 3 ways

In how many ways can you give 4 reports to only 2 secretaries? Choose the 2 secretaries in 3C2 ways. Each secretary must get at least one report so you can distribute them in 2 ways:
'1 and 3' - Choose one report in 4C1 ways and give it to one secretary in 2C1 ways. The other secretary gets the other 3 reports.
or
'2 and 2' - Choose 2 reports in 4C2 ways and give to the first secretary. The other secretary gets the other two reports.
Total number of ways is 3C2 * (4C1*2C1 + 4C2) = 42 ways

Note that in your method 2^4 is incorrect. It includes the ways in which all 4 reports go to one secretary.

So this means that in 3 + 42 = 45 ways, at least one secretary gets no report. In the rest of 81 - 45 = 36 ways, each secretary gets at least one report.

Required Probability = 36/81 = 12/27
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15 Jul 2006, 00:47
12
10
Total outcomes = 3^4 = 81

When each secretary is assigned atleast 1 report then two report will be assigned to 1 and 1 report each to remaining two.

Favourable outcomes = (Ways to select two report) * (Ways to select 1 seceratry) * (Ways to assign two report to two secretaries)
= (4C2) * (3C1) * (2!) = 36

Prob = 36/81
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Re: There are three secretaries who work for four departments.  [#permalink]

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17 Jul 2014, 20:34
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napoleonsselfie wrote:

The number of ways of ensuring at least one report goes to each secretary is 4C2 (select 2 reports out of 4 which go to the same person)*3C1 (select the person who must type 2 reports)*2! (since you have 2 reports left which you must distribute to the 2 remaining people such that each person gets one) = 36

Required probability = 36/81

Can someone explain why we must select the person who must type two reports (3C1)? That feels to me like double-counting, because in my mind, I feel like we already selected that person with 4C2. I'd like to know to understand completely so I don't continue to make mistakes on these kinds of problems. Thanks!

So here is the sequence of events you must follow:
Pick two reports out of 4 (R1, R2, R3, R4). You can do this in 4C2 ways such as (R1, R2) or (R1, R3) or (R2, R4) etc
Now go to the secretaries and select one secy from the 3 and hand over the two reports to her. You can do this in 3C1 ways such as A or B or C.
Now you have 2 reports and 2 secys. You can give them the reports in 2! ways.
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Re: There are three secretaries who work for four departments.  [#permalink]

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19 May 2014, 06:48
4
pretzel wrote:
How do we get the total outcomes as $$3^4$$= 81 ?

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.
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Re: There are three secretaries who work for four departments.  [#permalink]

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10 Dec 2013, 20:56
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jlgdr wrote:
mahesh004 wrote:
There are three secretaries who work for four departments. If each of the four departments have one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretary are assigned at least one report?

I got the same answer: Here's my reasoning

First Report you have 3 choices
Second Report you have 2 choices
Third Report you have 1 choice
Fourth report 3 choices again

Then total number of ways is: 3*2*1*3=3^2*2

This is not correct. You have assumed that the 4th report must go to someone who already has a report. There is no such constraint. You can easily give the 1st and 2nd reports to secretary 1, 3rd report to secretary 2 and 4th report to secretary 3. But you have ignored all such cases.

The number of ways of ensuring at least one report goes to each secretary is 4C2 (select 2 reports out of 4 which go to the same person)*3C1 (select the person who must type 2 reports)*2! (since you have 2 reports left which you must distribute to the 2 remaining people such that each person gets one) = 36

Required probability = 36/81
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15 Jul 2006, 09:54
1
mahesh004 wrote:

For each report there are three possible outcomes: Secretary A, Secretary B, Secretary C
Total outcome: 3^3=27
Total outcome that three secretary are assigned at least one report: P(3,3)=3!=6
Probability = 6/27=2/9

This is wrong solution. See this:

http://www.gmatclub.com/phpbb/viewtopic.php?t=14706
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Re: There are three secretaries who work for four departments.  [#permalink]

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18 May 2014, 08:25
1
What is wrong with..

Choose 3 out of 4 letters...4C3..assign them to each secretary...4C3 * 3! ways
Assign the remaining letter to any secretary: 3 ways

= 72/81
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Re: There are three secretaries who work for four departments.  [#permalink]

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Updated on: 18 May 2014, 19:17
1
JusTLucK04 wrote:
What is wrong with..

Choose 3 out of 4 letters...4C3..assign them to each secretary...4C3 * 3! ways
Assign the remaining letter to any secretary: 3 ways

= 72/81

Lets take an example to understand this. Lets say, we have 2 persons and 3 reports a, b,c. Then lets try and apply you logic:

3C2 =3. They can be given in 2! ways = 3*2 =6

1st 2nd
a b -------------------(1)
b c
c a
b a
c b ------------------(2)
a c

All right till here I suppose. Now the third report can be given in 2 ways, but you will see that there are repetitions.

(1) and (2) will produce the same result, ie ac,b and ac, b. Hence duplication. The problem with your approach is, you are trying to arrange all the reports, and therefore concerned about the fact the which report is given to him first, which is immaterial for our analysis.

Hope it works!!!

In fact, as you notice that a person getting two reports A and B can get it in 2 ways, A followed by B and otherwise, but it should be considered as a single case. Therefore, you need to divide 72 by 2 to get 36.

Originally posted by mittalg on 18 May 2014, 08:47.
Last edited by mittalg on 18 May 2014, 19:17, edited 1 time in total.
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Re: There are three secretaries who work for four departments.  [#permalink]

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18 May 2014, 21:24
1
VeritasPrepKarishma wrote:
JusTLucK04 wrote:
What is wrong with..

Choose 3 out of 4 letters...4C3..assign them to each secretary...4C3 * 3! ways
Assign the remaining letter to any secretary: 3 ways

= 72/81

"Reports R1, R2 and R3 are selected and given to secretaries A, B and C respectively and then report R4 is given to A", is a different case from "reports R2, R3 and R4 are selected and given to secretaries C, B and A respectively and then report R1 is given to A".

But actually both cases are same because A gets R1 and R4, B gets R2 and C gets R3. So there is double counting.

Ohh..Is there any foolproof specific approach for such problems so that we avoid such errors..Quite hard to catch them on the xam
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Re: There are three secretaries who work for four departments.  [#permalink]

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10 Jun 2015, 09:31
1
Departments : A,B,C,D
Secretaries : 1,2,3
As per the ques , one of the secretaries will have to type for 2 depts and the other two for 1 dept each.
Which means 2 dept. will give their report to 1 secy and the other two depts will give their report to the other 2 secys (1 each).
No. of ways to club the depts in the form of 2,1,1 = 6. (AB, C, D) (AC, B,D) (AD, B, C) (BC, A, D) (BD, A, C) (CD, A,B)
No. of ways of distributing the reports to 3 secys = 3!.
Therefore total no. of favorable cases = 6x3! = 36.
Total possible cases = 3^4 ( each dept can give its report to any of the secys).

Probability = 36/(3^4) = 4/9

Thanks
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Re: There are three secretaries who work for four departments.  [#permalink]

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27 Jul 2016, 22:42
1
suramya26 wrote:
why cant the total number of possiblities be
4^3?
I have used the following approach for it.
Kindly please tell me where i am wrong
assuming s1,s2,s3 be the secretaries and r1,r2,r3,r4 be the reports
if the reports are o be distributed among secretaries then
s1 can have 4 options of reports
similiarly s2,s3,s4 all will have 4 options
hence according to me 4x4x4=4^3=64 should be total ways
also please let me know the main concept of n^r takes place...

This concept cannot be used here.

Say there are 4 report options for s1.
s1 gets r1.

Now what about s2? Are there still 4 report options for her?

Also, each report has to be allotted to a secretary. Each secretary needn't be allotted a report. It is possible that s1 gets no report or gets 2 or 3 or even all 4 (when considering total number of cases).
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15 Jul 2006, 00:45
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Not pretty sure but think that the prob of any secretary typing a report 1/3 and NOT typing a reports is 2/3. Then the prob of NOT typing any of the reports is 16/81. We have 3 secretaries then 48/81 is the prob that one of them will not have a report to type. The required prob is 1-48/81=33/81=11/27
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15 Jul 2006, 01:40

For each report there are three possible outcomes: Secretary A, Secretary B, Secretary C
Total outcome: 3^3=27
Total outcome that three secretary are assigned at least one report: P(3,3)=3!=6
Probability = 6/27=2/9
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Re: There are three secretaries who work for four departments.  [#permalink]

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10 Dec 2013, 10:31
mahesh004 wrote:
There are three secretaries who work for four departments. If each of the four departments have one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretary are assigned at least one report?

I got the same answer: Here's my reasoning

First Report you have 3 choices
Second Report you have 2 choices
Third Report you have 1 choice
Fourth report 3 choices again

Then total number of ways is: 3*2*1*3=3^2*2

Total number of possibiliities = 3^4

So probability = 2/9

Cheers!
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10 Dec 2013, 20:49
mahesh004 wrote:

For each report there are three possible outcomes: Secretary A, Secretary B, Secretary C
Total outcome: 3^3=27
Total outcome that three secretary are assigned at least one report: P(3,3)=3!=6
Probability = 6/27=2/9

Here you are assuming there are 3 reports and 3 secretaries. But you are given that there are 4 reports (from 4 departments). Hence this is not correct.
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Re: There are three secretaries who work for four departments.  [#permalink]

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18 May 2014, 19:17
JusTLucK04 wrote:
What is wrong with..

Choose 3 out of 4 letters...4C3..assign them to each secretary...4C3 * 3! ways
Assign the remaining letter to any secretary: 3 ways

= 72/81

"Reports R1, R2 and R3 are selected and given to secretaries A, B and C respectively and then report R4 is given to A", is a different case from "reports R2, R3 and R4 are selected and given to secretaries C, B and A respectively and then report R1 is given to A".

But actually both cases are same because A gets R1 and R4, B gets R2 and C gets R3. So there is double counting.
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Re: There are three secretaries who work for four departments.  [#permalink]

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18 May 2014, 21:40
JusTLucK04 wrote:
VeritasPrepKarishma wrote:
JusTLucK04 wrote:
What is wrong with..

Choose 3 out of 4 letters...4C3..assign them to each secretary...4C3 * 3! ways
Assign the remaining letter to any secretary: 3 ways

= 72/81

"Reports R1, R2 and R3 are selected and given to secretaries A, B and C respectively and then report R4 is given to A", is a different case from "reports R2, R3 and R4 are selected and given to secretaries C, B and A respectively and then report R1 is given to A".

But actually both cases are same because A gets R1 and R4, B gets R2 and C gets R3. So there is double counting.

Ohh..Is there any foolproof specific approach for such problems so that we avoid such errors..Quite hard to catch them on the xam

Try to think about the solution in the way you will actually do it in practice. You should always try and assume a case to verify an answer as has been done by Karishma. I would say more practice will make you confident on these kinds of questions and enlighten you on the mistakes you may commit.
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Re: There are three secretaries who work for four departments.  [#permalink]

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19 May 2014, 01:26
JusTLucK04 wrote:

Ohh..Is there any foolproof specific approach for such problems so that we avoid such errors..Quite hard to catch them on the xam

Double counting is a very common mistake people new to P&C make - don't worry about it. Once you practice more, you will get used to taking a step back and ensuring that you are not double counting.
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Re: There are three secretaries who work for four departments.  [#permalink]

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19 May 2014, 06:27
How do we get the total outcomes as $$3^4$$= 81 ?
Re: There are three secretaries who work for four departments.   [#permalink] 19 May 2014, 06:27

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