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 Probability and Statistics Basic Probability and Statistics Math Forum

 June 7th, 2011, 02:00 AM #1 Member   Joined: May 2011 Posts: 51 Thanks: 0 Looking for better approaches-Probability 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+n-1)! 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+n-1)!/r!*n! = (r+n-1)!/r!*(n-1)! = (r+n-1)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 (n-1) ways and so on...= n! Then the total no. of ways are n!*r!/n!*(r-n)! = r!/(r-n)! After that I put the rest r-n elements and boxes in the sequence as above. Very similarly 1st I have to choose the box in n ways, then n-1 boxes and r-n elements can be suffled in (n-1+r-n)! = (r-1)!, so that is n*(r-1)! 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*(r-1)!*{r!/(r-n)!}*1/n!*r! = (r-1)!/(r-n)!*(n-1)! = (r-1)C(r-n) 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 approaches-Probability 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 r-n balls which must now be distributed into the n boxes. So use the first formula with r replaced by r-n June 7th, 2011, 07:11 PM #3 Member   Joined: May 2011 Posts: 51 Thanks: 0 Re: Looking for better approaches-Probability Thank you very much.  Tags approachesprobability Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post cacak Calculus 6 September 10th, 2013 06:30 AM MATHEMATICIAN Calculus 20 August 27th, 2013 05:50 PM bach71 Calculus 2 March 9th, 2013 12:50 PM rynoab Calculus 6 February 11th, 2013 07:50 PM strawberryPK Calculus 4 April 25th, 2011 05:12 PM

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