I tried using the equation listed in the book, but I ended up with 0 in the denominator, am I missing something?
Using the equation on page 378 I got 41.--Spfeifer 20:37, 11 February 2009 (UTC)
For this problem, I started looking at it as five indistinguishable objects into three indistinguishable boxes and then added the distinguishable part back in for the objects. You could have (5,0,0) (4,1,0) (3,2,0) (3,1,1) (2,2,1). Adding the distinguishable factor back in, for (5,0,0) there is 1 option, for (4,1,0) there is (5 choose 4)x(1 choose 1), for (3,2,0) there is (5 choose 3)x(2 choose 2), for (3,1,1) there is (5 choose 3)x(2 choose 1)x(1 choose 1), and for (2,2,1) there is (5 choose 2)x(3 choose 2)x(1 choose 1). Add these together and my answer was 66.
I'll verify that the response about the equation on page 378 is correct. As for getting a zero in the denominator, remember that 0! = 1. -NF
