Expert: Paul Klarreich - 10/16/2008

Question
A legal firm has 4 new cases and 3 lawyers.   How many ways can they assign 1 lawyer to each case if no lawyer works on more than 2 cases?

Answer
Questioner:   Adam
Category:  Calculus
 
Question:  A legal firm has 4 new cases and 3 lawyers.   How many ways can they assign 1 lawyer to each case if no lawyer works on more than 2 cases?
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Hi, Adam,

Try this:

You will partition the set {1,2,3,4} of cases into subsets each having 0-2 cases.  [You will not overlook the possibility that two lawyers get 2 each and the other gets none.]

List all the partitions.  Use some systematic method, such as:

First, put 1 with each of the others, and for each, partition the others.
{1-2}, {3}, {4}
{1-2}, {3-4}, {}
{1-3}, {2}, {4}
{1-3}, {2-4}, {}
{1-4}, {2}, {3}
{1-4}, {2-3}, {}

Then do that with 2, checking for duplicates.

{2-3}, {1}, {4}
{2-3}, {1-4}, {}  << delete
{2-4}, {1}, {3}
{2-4}, {1-3}, {} << delete.

Finally 3:

{3-4}, {1}, {2}
{3-4}, {1-2}, {}  << delete

Count the 'case partitions'.  Now, for each, count the ways they can be assigned to three lawyers, A,B,C.  That's a simple permutation problem.  

I think you can do the rest.