Math Forum :: View topic – Combinatorial problem

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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.

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