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yptsoi
Frequent VisitorJoined: 24 Jan 2004Posts: 46
Location: Hong Kong

Posted: Wed Feb 23, 2005 10:52 am Post subject: Combinatorial problem



n different balls are to be placed in r boxes (n>r>1) so that no box is left empty. How many arrangements can there be, if (i) the boxes are all different; (ii) the boxes are all identical.
Sorry for asking such simple question, but as my combinatorics skills are really too bad, I am not sure with my answer. Can anyone show me the correct answer? Thanks.





wch2eraq35
Frequent VisitorJoined: 04 Jan 2005Posts: 21
Location: HK,Tai Po

Posted: Wed Feb 23, 2005 10:56 pm Post subject:



When doing combination we may need them : when there are n objects , r of them are chosen number of combination is
or
where the P one , different order is counted different object
the C one , different order but same objects is counted one only





yptsoi
Frequent VisitorJoined: 24 Jan 2004Posts: 46
Location: Hong Kong

Posted: Thu Feb 24, 2005 9:26 pm Post subject:



……………….
so what is the answer of this question?





koopa
Frequent VisitorJoined: 21 Jan 2004Posts: 62

Posted: Fri Feb 25, 2005 2:58 pm Post subject:



For (ii) you can think about the problem as find the number of solutions to: x_1 + x_2 + … + x_r = n, where you view x_i as a box. The number of solutions is nC(r – 1). This is the answer to (ii) For (i) each permuted solution in (ii) is a solution thus the answer for (i) is r!*nC(r – 1).
Hope this helps.





