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There are three secretaries who work for four departments.

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

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New post 10 Jun 2015, 10:31
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

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

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New post 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 4C3 ways and distribute them among three secretaries in 3! ways => 4C3 * 3!
Then the remaining one report can be given to any of the three secretaries in 3 ways

All together => 4C3 * 3! * 3/3^4 => 8/9
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Re: There are three secretaries who work for four departments.  [#permalink]

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New post 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.  [#permalink]

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

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New post 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.  [#permalink]

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New post 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.  [#permalink]

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New post 27 Jul 2016, 23:42
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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Re: There are three secretaries who work for four departments.  [#permalink]

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New post 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.  [#permalink]

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New post 29 Jan 2017, 17:50
If distributing 4 balls among 3 persons so that all are distributed can be solved using the following formula:
\((n+r-1)C(r-1) = (4+3-1)C(3-1) = 6C2=15\), 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.  [#permalink]

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New post 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 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![/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 F1-S1;F2-S2;F3-S3;F4-S1.
Total Probability = ((1/3)^3)x(4!/2!) = 4/9
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Re: There are three secretaries who work for four departments.  [#permalink]

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New post 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.  [#permalink]

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New post 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. &nbs [#permalink] 28 Jun 2018, 11:06

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