Counting problem; Why is Instructor's solution correct?

How many ways can 9 identical DVDs be packed into 3 identical boxes if

(a) each box may contain any number of DVDs?

My instructor told me the following was correct: 1 box: {9}, 2 boxes: {8,1} {7,2} {6,3} {5,4}, 3 boxes: {7,1,1} { 6,2,1} {5,2,2} {4,3,2} {3,3,3}

1+4+5 = 10 ways.

My question is: Why can't 3 boxes have {3,1,5} or {4,1,4} ? Or is my instructor incorrect? Thanks.

Re: Counting problem; Why is Instructor's solution correct?

Quote:

Originally Posted by

**Walshy** How many ways can 9 identical DVDs be packed into 3 identical boxes if

(a) each box may contain any number of DVDs?

My instructor told me the following was correct: 1 box: {9}, 2 boxes: {8,1} {7,2} {6,3} {5,4}, 3 boxes: {7,1,1} { 6,2,1} {5,2,2} {4,3,2} {3,3,3}

1+4+5 = 10 ways.

My question is: Why can't 3 boxes have {3,1,5} or {4,1,4} ? Or is my instructor incorrect? Thanks.

You are correct the as is $\displaystyle 12$.

This is a well-known problem which takes many forms.

It is counting the number of ways a positive integer, $\displaystyle N$, can be partitioned into $\displaystyle k$ or fewer summands, P(N,k).

Example: $\displaystyle P(4,2)=3~:$ $\displaystyle 4,~3+1,~2+2$

$\displaystyle P(9,3)=12$.

Tell your instructor to read chapter six of __ MATHEMATICS OF CHOICE__ by Ivan Niven.