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Re: There are three secretaries who work for four departments.
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10 Jun 2015, 09:31
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, 03:45
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, 10:38
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, 02:32
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, 18:52
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, 22:29
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, 22:42
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, 10:27
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, 16:50
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, 07:39
[/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, 14:17
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, 10:06
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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