My Math Forum Looking for better approaches-Probability

 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: 706 Thanks: 141 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. $\binom{r+n-1}{r}$ 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 $\binom{(r-n)+n-1}{r-n}= \binom{r-1}{r-n}$
 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.

 Thread Tools Display Modes Linear 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

 Contact - Home - Forums - Cryptocurrency Forum - Top