5 different balls to 3 different people

mohish

New member
Joined
Mar 14, 2019
Messages
2
What is the total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets at least one ball?

The correct answer is 150 and I understand how it needs to be done.

However, my question is: Why can't we solve the question in the following way:

First we take any 3 balls out of 5 and distribute one ball each to the 3 persons. This can be done in C(5,3)*3! ways.

Now we are left with 2 remaining balls. So, we can do either of the following:
i) Give both those balls to one person. This, I believe, can be done in C(3,1) ways Or
ii) Give one ball each to two persons. This, I believe, can be done in C(3,2) ways

So, total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets at least one ball = C(5,3)*3! *[C(3,1) +C(3,2)] = 360

Why is wrong in the above approach? As I mentioned, the correct answer is 150.
 

mohish

New member
Joined
Mar 14, 2019
Messages
2
What is the total number of ways in which 5 balls of different colors can be distributed among 3 persons so that each person gets at least one ball?

The correct answer is 150 and I understand how it needs to be done.

However, my question is: Why can't we solve the question in the following way:

First we take any 3 balls out of 5 and distribute one ball each to the 3 persons. This can be done in C(5,3)*3! ways.

Now we are left with 2 remaining balls. So, we can do either of the following:
i) Give both those balls to one person. This, I believe, can be done in C(3,1) ways Or
ii) Give one ball each to two persons. This, I believe, can be done in C(3,2) ways

So, total number of ways in which 5 balls of different colors can be distributed among 3 persons so that each person gets at least one ball = C(5,3)*3! *[C(3,1) +C(3,2)] = 360

Why is wrong in the above approach? As I mentioned, the correct answer is 150.
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
3,036
Your method will double-count. For instance, the same distribution AB,CD,E could be attained by choosing ACE and distributing them, then putting B and D in their places, or first distributing BDE and then adding A and C, or several other ways. All these should only count as one; order doesn't count within one person.

What, as you understand it, is "how it needs to be done"? Probably there are several equally valid methods, and something different from what you were shown might be more satisfying to you.
 

pka

Elite Member
Joined
Jan 29, 2005
Messages
7,813
What is the total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets at least one ball?
We are not prepared to offer tutorials here. So I can give you only brief outline. Here is a page relative to this problem.
This question is the equivalent of asking "how many surjections(onto functions) are there from a set of five to a set of three.
In general if \(\displaystyle \|A\|=m>\|B\|=n\) the number of onto functions \(\displaystyle A\to B\) is \(\displaystyle \sum\limits_{k = 0}^n {{{( - 1)}^k}\binom{ n} { k} {{(n - k)}^m}}\)
In this case \(\displaystyle n=3\;\&\;m=5\) SEE HERE
 
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