
Probability and Statistics Basic Probability and Statistics Math Forum 
 LinkBack  Thread Tools  Display Modes 
May 3rd, 2017, 11:35 PM  #1 
Newbie Joined: May 2017 From: phoenix, az Posts: 7 Thanks: 0  counting
I have 3 distinguishable balls and two boxes labeled box 1 and box 2. How many different ways can I put the 3 balls into the 2 boxes? Thanks 
May 4th, 2017, 12:00 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,424 Thanks: 1311 
you can choose 2 balls from the three in $\binom{3}{2}=3$ ways and then there are two ways to arrange the 2 balls into the 2 boxes so there are a total of $2 \cdot 3 = 6$ arrangements 
May 4th, 2017, 01:07 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,613 Thanks: 2071 
That's incorrect. However, the question is unclear as to whether the ways to be counted include putting all three balls into the same box.

May 4th, 2017, 04:33 AM  #4  
Senior Member Joined: Feb 2010 Posts: 706 Thanks: 140  Quote:
If empty boxes are allowed, then the answer is 8. If every box must contain at least one ball, then the answer is 6.  
May 4th, 2017, 09:02 PM  #5 
Newbie Joined: May 2017 From: phoenix, az Posts: 7 Thanks: 0 
I see that if you can only have 1 ball in each box, but you must have a ball in both boxes, then the answer is 6 ({1,2},{1,3},{2,1},{2,3},{3,1},{3,2}), but if you can have a box that is empty, the answer is 12 ({1,2},{1,3},{2,1},{2,3},{3,1},{3,2},{1,0},{0,1},{ 2,0},{0,2},{3,0},{0,3}). If you can have two balls in the same box, but no empty boxes then the answer is 12. If you can have 2 or 3 balls in the same box, therefore having an empty box, then the answer is 20. So you're right, the question is very ambiguous.

May 5th, 2017, 02:54 AM  #6  
Senior Member Joined: Feb 2010 Posts: 706 Thanks: 140  Quote:
Does this mean you put one ball in the left box and two in the right box? If so, then you haven't said which number ball goes in the left box? Or perhaps it means you put ball #1 in the left box, ball #2 in the right box and you toss ball 3 in the back yard for your dog to play with. If you must have at least one ball in each box then the solution is 6 (boxes are numbered 1 and 2, balls are numbered 1,2,3): box 1 box 2 1  23 2  13 3  12 23  1 13  2 12  3 If you allow empty boxes then you have the same six plus these two: box 1 box 2 123   123  
May 5th, 2017, 02:54 PM  #7 
Newbie Joined: May 2017 From: phoenix, az Posts: 7 Thanks: 0 
{1,2} means ball #1 in the left box (or box #1), ball #2 in the right box (or box #2), and you leave the third ball out for the dog to play with as you say. You're assertion is correct only under a limited amount of options, otherwise you are wrong.

May 5th, 2017, 03:02 PM  #8  
Senior Member Joined: Feb 2010 Posts: 706 Thanks: 140  Quote:
Make up your mind.  
May 5th, 2017, 03:20 PM  #9 
Senior Member Joined: Feb 2010 Posts: 706 Thanks: 140 
Let me rephrase the question with my own assumptions (if you don't like them, tell me what you want): I have two boxes labelled 1 and 2. I have three balls labelled 1, 2, and 3. In how many ways can I put the 3 balls into the 2 boxes under the following assumptions: 1. I don't have to put the 3 balls into the 2 boxes. 2. (for purposes of clarity) a box may be empty.The answer is 27. Last edited by mrtwhs; May 5th, 2017 at 03:25 PM. 
May 5th, 2017, 07:27 PM  #10 
Newbie Joined: May 2017 From: phoenix, az Posts: 7 Thanks: 0 
This is just the way that the question was presented to me on a test, I answered six and got the problem wrong. I think we can both agree that the question is incomplete.


Tags 
counting, probability 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Counting  USAMO Reaper  Probability and Statistics  2  February 9th, 2015 01:21 PM 
Counting  jbergin  Advanced Statistics  1  December 28th, 2014 03:02 PM 
Counting  azizlwl  Algebra  4  February 16th, 2012 04:53 PM 
counting  soulsister5  Applied Math  5  May 10th, 2010 02:44 PM 
More Counting  roguebyte  Applied Math  0  December 31st, 1969 04:00 PM 