
Algebra PreAlgebra and Basic Algebra Math Forum 
 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. 
March 15th, 2019, 12:15 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,531 Thanks: 1390 
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. 
March 15th, 2019, 04:41 AM  #3  
Global Moderator Joined: Dec 2006 Posts: 20,939 Thanks: 2210  Quote:
Quote:
Quote:
That should be C(5,3)*3! *[C(3,1)/3 + 2C(3,2)/4] = 150.  
March 15th, 2019, 06:27 AM  #4 
Newbie Joined: Mar 2019 From: bangalore Posts: 7 Thanks: 0  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. 
March 15th, 2019, 01:11 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,939 Thanks: 2210 
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.

March 20th, 2019, 05:07 AM  #6 
Newbie Joined: Mar 2019 From: bangalore Posts: 7 Thanks: 0  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. 
March 20th, 2019, 08:39 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 20,939 Thanks: 2210 
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. 
July 6th, 2019, 04:16 AM  #8 
Newbie Joined: Jun 2016 From: Hong Kong Posts: 25 Thanks: 2 
The general way is Stirling numbers of the second kind $3!\left\{ {5\atop 3} \right\}=3^53\times 2^5+3\times 1=150$ 

Tags 
balls, people 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
nballs  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 