On one side of a river are three huminds, one big monkee, two small monkees and one boat. 
Each of the huminds and the big monkee are strong enough to row the boat. The boat can fit 
one or two bodees (regardless of size). If at any time at either side of the river the monkees 
outnumbah the huminds, the monkeys will eat the huminds. (Be sure to include those in the 
boat with those on the shore when you do your counting.) How do you get everyone on the 
other side of the river alive?  

This is the solution that came with the puzzle.  The letters in parentheses represent either side 
of the river.  s=small monkey, b=big monkey.

1.  Row b and s over ( hhhs / bs )
2.  Row b back ( hhhbs / s )
3.  Row b and s over ( hhh / bss )
4.  Row b back ( hhhb / ss )
5.  Row h and h over ( hb / hhss )
6.  Row h and s back ( hhbs / hs )
7.  Row h and b over ( hs / hhbs )
8.  Row h and s back ( hhss / hb )
9.  Row h and h over ( ss / hhhb )
10. Row b back ( bss / hhh )
11. Row b and s over ( s / hhhbs )
12. Row b back ( bs / hhhs )
13. Row b and s over (  / hhhbss )
  
Using the format your solution uses, my "solution" would be written

1.  Row b and s over ( hhhs / bs )
2.  Row b back ( hhhbs / s )
3.  Row b and h over ( hhs / bhs ) tilt! (too many monkees - humind is history!)

When they say "at either side of the river" they mean you've got to count those in the boat 
*and* those on the shore the boat is near. My "solution" was assuming you only had to worry 
about who was on the shores. I think this technically occured to me but I blew it off and 
decided the huminds didn't need to worry about getting eaten during the transition periods. Silly 
me. My "solution" is wrong. Hey, I'm only humind. Ta err is humind.

Nice problem.