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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.

mmcrjx
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NinjaWarrrior
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1 Answers1

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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$(11+11+3=25)$.

mmcrjx
  • 654