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Re: There are three secretaries who work for four departments.
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10 Jun 2015, 10:31
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 Rajat



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Re: There are three secretaries who work for four departments.
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12 Jun 2016, 04:45
ps_dahiya wrote: 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 Hi, Please explain the mistake in my approach. Each of the reports can be given to any of the three secretaries. so, each report has 3 options => 3^4 Now, to make sure that each secretary gets atleast one, I select 3 reports from 4 reports available in 4 C3 ways and distribute them among three secretaries in 3! ways => 4 C3 * 3! Then the remaining one report can be given to any of the three secretaries in 3 ways All together => 4 C3 * 3! * 3/3^4 => 8/9



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Re: There are three secretaries who work for four departments.
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28 Jun 2016, 11:38
Great Question!! Required 7:48 min to solve. Here's my approach
Total possibilities : 3^4=81
Now, I have 4 reports and 3 persons. I can disbribute these reports in the following ways: A B C 1 1 2 1 2 1 2 1 1 which comes out to 3 possibilities
So, Total ways of allocating these 4 reports = 4!/2 * 3 possibilites
Therefore Prob = 4*3*3 / 81 = 4/9



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Re: There are three secretaries who work for four departments.
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11 Jul 2016, 03:32
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?
A. 8/9 B. 64/81 C. 4/9 D. 16/81 E. 5/9 Quote: Hi Bunuel, Can you please provide a detailed explanation? Thanks in advance.
Regards, Yosita



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Re: There are three secretaries who work for four departments.
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11 Jul 2016, 19:52
The number of ways to divide 4 reports in 3 lots s.t. every secretory gets at least one report is (1,1,2). Distributed lots are given by
= 4!x 3!/(1! x 1! x 2! x 2!). Total outcomes = 3^4. Probability = 4/9



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Re: There are three secretaries who work for four departments.
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27 Jul 2016, 23:29
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...



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Re: There are three secretaries who work for four departments.
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27 Jul 2016, 23:42
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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Re: There are three secretaries who work for four departments.
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30 Jul 2016, 11:27
Bunuel wrote: 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. Hey Bunnel, Why not the other way that each Secretary has 4 options so 4*4*4 ? I always get confused in such questions. Are there are any similar questions for practice, around this concept ? Thank you!



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There are three secretaries who work for four departments.
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29 Jan 2017, 17:50
If distributing 4 balls among 3 persons so that all are distributed can be solved using the following formula: \((n+r1)C(r1) = (4+31)C(31) = 6C2=15\), why we can't solve it using the same formula? Isn't it like distributing 4 reports among 3 secretaries so that all reports are distributed?
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Re: There are three secretaries who work for four departments.
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27 Jun 2018, 08:39
[/quote]
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[/quote]
Can someone explain why we must select the person who must type two reports (3C1)? That feels to me like doublecounting, 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![/quote]
I have alternate and easier solution : There are four files and secretaries have to be assigned to every file. Files are  F1,F2,F3,F4 ; Secretaries are : S1,S2,S3. F1 can get any secretary independently. So P (F1 getting S1) = 1/3 P (F2 getting S2) = 1/3 P (F3 getting S3) = 1/3 After this we dont care what F4 gets because all Secretaries has got one file. Suppose F1S1;F2S2;F3S3;F4S1. Total Probability = ((1/3)^3)x(4!/2!) = 4/9



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Re: There are three secretaries who work for four departments.
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27 Jun 2018, 15:17
I spent a lot of time trying to make sense of this question. I got the answer but I'm not sure if it's the correct way or if I will even remember considering this approach.
Total possibility = 3^4 = 81
A= secretary 1 B= Secretary 2 C = Secretary 3
4 reports needs to be spread out amongst the 3 secretaries where at least 1 report is completed can happen in the following ways:
(1) AABC (2) ABBC (3) ABCC
(1) AABC 4C2 = 6 > 4 reports but choose 2 for A 2C1 = 2 reports but choose 1 for B 1C1 = 1 report but choose 1 for C
4C2 * 2C1 * 1C1 = 12
(2) ABBC 4C2 = 6 > 4 reports but choose 2 for B 2C1 = 2 reports but choose 1 for A 1C1 = 1 report but choose 1 for C
4C2 * 2C1 * 1C1 = 12
(3) ABCC 4C2 = 6 > 4 reports but choose 2 for C 2C1 = 2 reports but choose 1 for A 1C1 = 1 report but choose 1 for B
4C2 * 2C1 * 1C1 = 12
12+12+12 = 36
36/81 = 4/9



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Re: There are three secretaries who work for four departments.
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28 Jun 2018, 11:06
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?
A. 8/9 B. 64/81 C. 4/9 D. 16/81 E. 5/9 Total # of ways of assigning a secretary for each report = 3*3*3*3 = 81 ways # of ways in which each secretary is assigned at least one report can be done by first assigning one report to each secretary & then choosing one secretary out of the three for the fourth report = # of ways of assigning the reports as (1, 1, 2) /(1, 2, 1) / (2, 1, 1) = 4*3*1*3!/2! = 12 * 3 = 36 Hence required probability = 36/81 = 4/9 Answer C. Thanks, GyM
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Re: There are three secretaries who work for four departments.
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29 Mar 2019, 07:53
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?
A. 8/9 B. 64/81 C. 4/9 D. 16/81 E. 5/9 Isn't this too difficult a question to come in GMAT? #justasking



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Re: There are three secretaries who work for four departments.
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29 Mar 2019, 22:59
Dear VeritasKarishma , I am really confused ,can you please let me know where's my mistake : total ways : 3^4 now , let's choose 3 reports in 4C3 and give them to the three secretaries in 3! ways : we are left with one report : it hase 3 choices to go (each for a secretary) so the wanted ways : 4C3 * 3! * 3 = 4 *3*2* 3 so I'm getting 8/9 in another approach : I'm getting the correct answer : let's choose 2 reports 4C2 and give them to one secretary *3 , then the 2 left reports will go to the 2 left secretaries in 2 ways so I get : 4C2 *3 *2 =2 *3 *3 *2 so p = 4/9 can you please let me know why my first is wrong Thanks
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Re: There are three secretaries who work for four departments.
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30 Mar 2019, 23:54
foryearss wrote: Dear VeritasKarishma , I am really confused ,can you please let me know where's my mistake : total ways : 3^4 now , let's choose 3 reports in 4C3 and give them to the three secretaries in 3! ways : we are left with one report : it hase 3 choices to go (each for a secretary) so the wanted ways : 4C3 * 3! * 3 = 4 *3*2* 3 so I'm getting 8/9 in another approach : I'm getting the correct answer : let's choose 2 reports 4C2 and give them to one secretary *3 , then the 2 left reports will go to the 2 left secretaries in 2 ways so I get : 4C2 *3 *2 =2 *3 *3 *2 so p = 4/9 can you please let me know why my first is wrong Thanks There is double counting in your method 1: Say there are 4 reports R1, R2, R3, R4 and 3 secys S1, S2, S3 You select 3 of the 4 reports (R1, R2 and R3) and give them to S1, S2 and S3 respectively. Then you give R4 to S1. or You select 3 of the 4 reports (R4, R2 and R3) and give them to S1, S2 and S3 respectively. Then you give R1 to S1. In both cases, S1 has R1 and R4, S2 has R2 and S3 has R3. But you counted them as two different cases. Hence this approach is not correct.
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Re: There are three secretaries who work for four departments.
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31 Mar 2019, 00:04
VeritasKarishma wrote: foryearss wrote: Dear VeritasKarishma , I am really confused ,can you please let me know where's my mistake : total ways : 3^4 now , let's choose 3 reports in 4C3 and give them to the three secretaries in 3! ways : we are left with one report : it hase 3 choices to go (each for a secretary) so the wanted ways : 4C3 * 3! * 3 = 4 *3*2* 3 so I'm getting 8/9 in another approach : I'm getting the correct answer : let's choose 2 reports 4C2 and give them to one secretary *3 , then the 2 left reports will go to the 2 left secretaries in 2 ways so I get : 4C2 *3 *2 =2 *3 *3 *2 so p = 4/9 can you please let me know why my first is wrong Thanks There is double counting in your method 1: Say there are 4 reports R1, R2, R3, R4 and 3 secys S1, S2, S3 You select 3 of the 4 reports (R1, R2 and R3) and give them to S1, S2 and S3 respectively. Then you give R4 to S1. or You select 3 of the 4 reports (R4, R2 and R3) and give them to S1, S2 and S3 respectively. Then you give R1 to S1. In both cases, S1 has R1 and R4, S2 has R2 and S3 has R3. But you counted them as two different cases. Hence this approach is not correct. Thank you for the explanation , that makes sense , is there any way how to find hidden double counting ? I mean both methods seems logical unless I spend descent time testing some possible values like you did :\
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Re: There are three secretaries who work for four departments.
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