How many ways are there to distribute 25 identical balls among 5 players where each player must get at least 1 and no player may get 10 or more? We know that
$$x_1 + x_2 + x_3 + x_4 + x_5 = 25;  1\leq x_i < 10$$.
Therefore, I'm thinking about getting all possible combinations and subtracting them by where 4 people get at least 1 and one other gets at least 11 so we'd have 
$$x_1 + x_2 + x_3 + x_4 + x_5 = 11$$
Getting all the possible combinations (multiplying by 5 since any one of them could have more than 10 balls)
And subtract both of the answers.
 A: An intuitive way to think about this is to consider a problem of arranging 0's and 1's.
Say you have n balls to distributed amongst k people with the same conditions as above. That each person should get atleast $1$ ball and the number of balls with one person can be at max $l$. Therefore the problem is as follows:
You have $n$ 1's and $k-1$ zeros to arrange in such a way that the zeros don't lie in the ends and zeros are not adjacent. This ensures the minimum 1 ball per person condition. The sum of the of the 1's between zeros is the number of balls for that person.
Example:
$$111011110110\cdots$$
Therefore $x_1=3$, $x_2=4$, $x_3=2$  and so on. There are $n-1$ holes to be filled amongst the $n$ 1's by $k-1$ zeros. And this can be done in ${n-1}\choose{k-1}$ ways.
And about each person having a maximum of $l-1$ balls condition can be enforced by subtracting the cases in which one person has $l$ or more balls. This is easy in this case because only 2 people can have 11 balls at a time$(11+11+3=25)$.
