Pls can any one explain the concept as how to solve this Maths question

[rschunti]

In how many ways 4 files A,B,C,D can be distributed to 3 typists so that each typist gets at least one file to type?(Note:- Any file can be given to any typist).

[DWarrior]

You're distributing 3 files and then the last file gets assigned to someone.

First, get the number of ways 3 files can be distributed: 4P3. You're using permutations because it's important which typist gets which file.

Then, for every permutation, there are 3 ways to distribute the last file, so multiply 4P3 by 3.

Finally, don't forget that this will double count each entry. For example, if 1 gets A, 2 gets B, 3 gets C, and 1 gets D, that's the same as 1 getting D, 2 getting B, 3 getting C, and 1 getting A. Every way to distribute the last file will have flipped version where that file could have been assigned as part of the permutation. To remove double counting, just divide by two.

The equation should be: 4P3 * 3/2, 36 different ways.

[kevinspain]

Originally Posted by rschunti

In how many ways 4 files A,B,C,D can be distributed to 3 typists so that each typist gets at least one file to type?(Note:- Any file can be given to any typist).

Step One: Choose 2 of 4 files that will be assigned to one person
4C2= 4(3)/2=6 ways

Step Two: Choose 'lucky' typist
3 ways

Step Three: Distribute two remaining files to the two remaining typists
2 ways

Product 36

[prreeet]

Originally Posted by kevinspain

Step One: Choose 2 of 4 files that will be assigned to one person
4C2= 4(3)/2=6 ways

Step Two: Choose 'lucky' typist
3 ways

Step Three: Distribute two remaining files to the two remaining typists
2 ways

Product 36

Nice one...i agree

[GMAD]

Originally Posted by kevinspain

Step One: Choose 2 of 4 files that will be assigned to one person
4C2= 4(3)/2=6 ways

Step Two: Choose 'lucky' typist
3 ways

Step Three: Distribute two remaining files to the two remaining typists
2 ways

Product 36

Great explanation. This really does simplify the problem. Thanks!

[ANSHU1]

In a certain company, three typists service for four departments. If the departments each send a file to the typists at random, what is the probability that every typist will recieve at least one file?

A. 8/9
B. 64/81
C. 4/9
D. 16/81
E. 5/9


There are two convincing answers for this question and I'm finding it difficult to select one. Can you please explain.


Soution - I

Total outcomes = 3 * 3 * 3 * 3

Each file can be sent to any of the three typists. So file 1 can be sent to typist 1,2 or 3. Similarly for the rest of the files.

Favourable Outcomes = 4C3 * 3! *3

First, we need to select 3 files out of 4 and distribute them to each of the three typists so that each one of them has one file. So we select the files in 4C3 ways and multiply it by 3! because 3 typists are involved and each of the three could get the 3 files in 3! ways. (So as typist 1, I could get file no 1, 2 or 3; As typist 2 I can get any of the files, excepting the one that typist 1 has already got; As typist 3 I can get only 1 file - the file that neither typist 1 or 2 have got. Hence 3 * 2 *1 or 3!)

Now, we have one file remaining of the four. This file has to be given to one of the three typists. Since it could be any of the three typists, we can say that that one file can be distributed in 3 ways.

Probability = Favourable outcomes/ Total outcomes

= (4 * 3 * 2 *3)/ (3 * 3 * 3 * 3)

= 8/9



Solution - II

Three typists A, B, and C.

In how many ways can one or two typists receive no file?

(1). A or B or C received all four files, 3 ways;

(2). A received no file, then 4 files will be distributed among B amd C:

2*2*2*2=16, just same as 3^4=81

The 16 ways include C=4, B=0 or C=0, B=4, which were counted in step (1). so, 16-2=14 favorable ways;

Same way, when B received no file: 14

C received no file: 14

Totally, in 3+14*3=45 ways, one or more typists will be leisure.

Therefore, 81-45=36 ways can satisfy the condtions.

36/81=4/9


KEVINSPAN ANSWES THIS QUESTION WELL.

[bose]

good one. i thot it was 72 forgot to divide by 2 to remove duplicates