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June 7th, 2011, 02:00 AM  #1 
Member Joined: May 2011 Posts: 51 Thanks: 0  Looking for better approachesProbability
Q: There are n distinct boxes in which I have to put r identical elements. How many ways are there? What if no boxes would be empety? A: First we have to chose the box then the elements. If we will write them then it will appear as a sequence, but any sequence like this has to start with any box. So first box will be chosen in n ways then the rest boxes and elements can be suffeled in (r+n1)! ways. The elements are identical so can be altered in r! ways, also the boxes with elements in it can be altered in n! ways So the answer is n*(r+n1)!/r!*n! = (r+n1)!/r!*(n1)! = (r+n1)Cr Now the 2nd answer... We don't want any box empety. First I took all identical elements and wrote 1to r on each of them, so now they became distinct. Then I choose n elements among those in rCn ways. Now I will put 1 element in each box...so for 1st box I will chose 1 element in n ways, then for the next (n1) ways and so on...= n! Then the total no. of ways are n!*r!/n!*(rn)! = r!/(rn)! After that I put the rest rn elements and boxes in the sequence as above. Very similarly 1st I have to choose the box in n ways, then n1 boxes and rn elements can be suffled in (n1+rn)! = (r1)!, so that is n*(r1)! If we will remove those marks then r identical elements can be altered in r! ways, also the boxes with elements in it can be altered in n! ways It brings us...n*(r1)!*{r!/(rn)!}*1/n!*r! = (r1)!/(rn)!*(n1)! = (r1)C(rn) The answers are right but for the second question it seems I have taken a long way, am I missing something? Is there a better and direct approach? 
June 7th, 2011, 11:22 AM  #2 
Senior Member Joined: Feb 2010 Posts: 711 Thanks: 147  Re: Looking for better approachesProbability
You have n distinct boxes and r identical ping pong balls. For your first question, I agree with your answer. For your second question, here is a different way to get your answer. First put one ping pong ball into each box. This leaves you with rn balls which must now be distributed into the n boxes. So use the first formula with r replaced by rn 
June 7th, 2011, 07:11 PM  #3 
Member Joined: May 2011 Posts: 51 Thanks: 0  Re: Looking for better approachesProbability
Thank you very much. 

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