Thread: 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.

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 .
This is a well-known problem which takes many forms.
It is counting the number of ways a positive integer, , can be partitioned into or fewer summands, P(N,k).

Example:

.

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