My Math Forum  

Go Back   My Math Forum > High School Math Forum > Algebra

Algebra Pre-Algebra and Basic Algebra Math Forum


Thanks Tree4Thanks
  • 1 Post By romsek
  • 1 Post By skipjack
  • 1 Post By skipjack
  • 1 Post By skipjack
Reply
 
LinkBack Thread Tools Display Modes
March 14th, 2019, 10:49 PM   #1
Newbie
 
Joined: Mar 2019
From: bangalore

Posts: 7
Thanks: 0

5 different balls to 3 different people

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.

Last edited by skipjack; March 15th, 2019 at 02:57 AM.
mohish is offline  
 
March 15th, 2019, 12:15 AM   #2
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: USA

Posts: 2,373
Thanks: 1276

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.
Thanks from mohish
romsek is online now  
March 15th, 2019, 04:41 AM   #3
Global Moderator
 
Joined: Dec 2006

Posts: 20,373
Thanks: 2010

Quote:
Originally Posted by mohish View Post
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.

Quote:
Originally Posted by mohish View Post
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.

Quote:
Originally Posted by mohish View Post
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.

Quote:
Originally Posted by mohish View Post
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.
Thanks from mohish
skipjack is offline  
March 15th, 2019, 06:27 AM   #4
Newbie
 
Joined: Mar 2019
From: bangalore

Posts: 7
Thanks: 0

Quote:
Originally Posted by skipjack View Post
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?

Last edited by mohish; March 15th, 2019 at 06:37 AM.
mohish is offline  
March 15th, 2019, 01:11 PM   #5
Global Moderator
 
Joined: Dec 2006

Posts: 20,373
Thanks: 2010

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.
Thanks from mohish
skipjack is offline  
March 20th, 2019, 05:07 AM   #6
Newbie
 
Joined: Mar 2019
From: bangalore

Posts: 7
Thanks: 0

Quote:
Originally Posted by skipjack View Post
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.
mohish is offline  
March 20th, 2019, 08:39 AM   #7
Global Moderator
 
Joined: Dec 2006

Posts: 20,373
Thanks: 2010

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.
Thanks from mohish
skipjack is offline  
Reply

  My Math Forum > High School Math Forum > Algebra

Tags
balls, people



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
n-balls Adam Ledger Number Theory 6 May 2nd, 2016 11:38 AM
Select 5 balls from 12 balls, where some balls are identical Skyer Algebra 3 January 6th, 2014 11:26 AM
How many balls can fit into a jar daigo Algebra 2 July 8th, 2012 09:47 AM
20 balls for 4 people tiba Algebra 3 June 20th, 2012 04:29 AM
An urn contains a azure balls and c carmine balls.. (Help) byron123 Advanced Statistics 2 September 10th, 2008 09:39 AM





Copyright © 2019 My Math Forum. All rights reserved.