12 identical balls and 3 teams I was wondering for the following question:
There’s a box with 12 identical balls. 
1) In how many ways 3 teams can pick them given that each can hold a maximum of 10 balls.
2) How many ways each team can pick them given that each team can pick a maximum of 10 balls and a minimum of 1 ball?
A step by step explanation will be helpful. Thanks
 A: It is easier to find it by contradiction: 
1) What is the total number to distribute the balls among the teams? (stars-and-bars)
2) In how many ways a team can have more than 10? 
3) You have 3 teams.
4) Subtract this number from $\binom{14}{2}$ to get what you need. 
A: ooooo|oooo|ooo
Could be one representation of the allocation between balls between teams.
i.e. 5 balls to team 1, 4 balls to team 2, 3 balls to team 3.
How many ways can we allocate circles and pipes (or more traditionally "stars and bars")?
There are 2 pipes and 14 objects total  ${14\choose2}$
So this is all of the ways to allocate 12 balls.
How many ways to allocate 10 balls such that no one has more than 10?
Lets answer this question instead: How many ways to allocate 10 balls such that someoene has at least 10?
Give 10 to 1 team allocate the rest.
There are 2 balls to distribute (and 2 pipes) ${4\choose2}$
And there are 3 teams that could be the "rich team"
$3\cdot{4\choose2}$
Since there is a question here. Lets enumerate the cases.  I say there are 18 
cases.
(10,2,0), (10,1,1), (10,0,2), (11,1,0), (11,0,1), (12,0,0) (that is 6 so far),
(2,10,0), (1,10,1),(0,10,2), (1,11,0),(0,11,1), (0,12,0) (another 6)
and 6 when team 3 gets the big pile of balls.
How many ways to allocate 10 balls such that no one has more than 10?
${14\choose2}-3\cdot{4\choose2}$
How many ways to allocate 10 balls such that no one has more than 10, and every team has at least one?
It is impossible for a team to have more than 10 balls, if every team has at least one.
Give one ball to each team.  Allocate the rest
that leaves 9 balls to distribute (and 2 pipes)
${11\choose2}$
