5 different balls to 3 different people

Mar 2019
7
0
bangalore
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

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

Math Team
Sep 2015
2,960
1,674
USA
I'd look at it this way.

There are two basic arrangements

1,2,2 and 1,1,3

There are $\dbinom{3}{2}$ ways to choose the 2 people that get 2 balls.
Then we choose a ball from the 5 and give it to the person that gets 1 ball.
We choose 2 balls from the remaining four and give it to one of the people that gets 2 balls. The 3rd person gets the remaining balls.

So we have

$\dbinom{3}{2}\dbinom{5}{1}\dbinom{4}{2} = 3 \cdot 5 \cdot 6 = 90$ ways to arrange the 5 different balls among 3 different people where 2 people get 2 balls.

There are $\dbinom{3}{1}$ ways to choose the person that gets 3 balls.
We choose 1 ball from 5 and give it to a person that gets 1 ball.
We choose 1 ball from the remaining 4 and give it to the other.
The third person gets the remaining 3 balls.

So we have

$\dbinom{3}{1}\dbinom{5}{1}\dbinom{4}{1} = 3 \cdot 5 \cdot 4 = 60$ ways to arrange the balls where 1 person gets 3 balls.

$90+60=150$

Note: The order that we distribute the balls doesn't matter. All the orders will result in the same numbers. Try it and see.
 
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skipjack

Forum Staff
Dec 2006
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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.
Correct.

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
Correct, but that person's result is achievable in C(3,1) = 3 ways.

Or
ii) Give one ball each to two persons. This, I believe, can be done in C(3,2) ways
Incorrect - it can be done in C(3,2)*2! ways, but the same results are achievable in 2!*2! = 4 ways.

So, total . . . = C(5,3)*3! *[C(3,1) +C(3,2)] = 360 What is wrong?
That should be C(5,3)*3! *[C(3,1)/3 + 2C(3,2)/4] = 150.
 
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Mar 2019
7
0
bangalore
Correct, but that person's result is achievable in C(3,1) = 3 ways.
Thanks for your reply. Can you please explain this and why do you have this in the denominator in the final equation?
 
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skipjack

Forum Staff
Dec 2006
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If green and red are received by the person who already has blue, you get the same result as when that person already had green and then receives red and blue. As you count the same result three times, your count needs to be divided by three.
 
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Mar 2019
7
0
bangalore
the same results are achievable in 2!*2! = 4 ways.
Hi @skipjack, can you please explain how you arrived at the 2!*2! figure? I do understand conceptually that you've divided the final equation with 2!*2! to avoid repetitions.

But what is the exact Mathematical formula behind this? Would really help me get a deeper understanding.
 

skipjack

Forum Staff
Dec 2006
21,478
2,470
A particular result, such as person 1 receives red and green, whilst person 2 receives blue and yellow is achievable in C(2,1) (or P(2,2)) ways, which is 2 (or 2!), for each person.

In the above example, person 1 can receive red first or green first, and person 2 can receive blue first or yellow first. That's 2*2 = 4 ways in total, which you had counted as four results, rather than just one.

The general formula you wanted would depend on how the problem is generalized.
 
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