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

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!

[/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

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
_________________

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

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 :\