5,0,0
0,5,0
0,0,5

1,1,3
1,3,1
3,1,1

1,2,2
2,1,2
2,2,1

3,2,0
2,3,0
0,3,2
0,2,3
3,0,2
2,0,3

0,1,4
0,4,1
4,1,0
1,4,0
4,0,1
1,0,4


21

There's another way of doing this without writing all the answer options..

Just write the different ways you can distribute 5 donuts..

a. 1, 2, 2 can be arranged in 3!/2! ways
b. 3, 2, 0 can be arranged in 3! ways
c. 4, 1, 0 can be arranged in 3! ways
d. 5, 0, 0 can be arranged in 3!/2! ways
e. 1, 1, 3 can be arranged in 3!/2! ways

Since we have to distibute donuts.. 0, 0, 0 is not an option, although 0 is a whole number.

Total ways = a + b + c + d + e = 21

Brent,

Isn't this similar to one of the problems you created?? I remember you created one and there was a trick to it, but I don't quite remember what it was.

Brent Hanneson wrote:

There it is!! I knew it was something like that but couldn't quite remember!

Brent Hanneson wrote:

Damn cool man!! Nice approach.. I liked it a lot.. super..