There are three secretaries who work for four departments.

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.

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.

How do we get the total outcomes as \(3^4\)= 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!

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

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

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

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

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...

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!

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?
_________________

[/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

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

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
_________________

Isn't this too difficult a question to come in GMAT?
#justasking

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 :\