# How many ways there are?

I cant solve the following problem. In how many ways we can divide 6 balls between 3 children if every children must receive at least 1 ball. I don't understand the problem. Is it permutations or what?

• Are the balls the same or different? – Ofir Schnabel Jan 26 '15 at 10:29

If the balls are identical, first give each child one ball. Then lay the three remaining balls in a row with two "dividers". How many ways can you arrange these three balls and two dividers?

We can use the stars and bars (see here) to count the ways we can divide $6$ balls between $3$ children if every children must receive at least 1 ball.

For any pair of positive integers $n$ and $k$, the number of distinct $k$-tuples of positive integers whose sum is $n$ is given by the binomial coefficient $${n-1\choose k-1}.$$

Hence, the answer to the question is $${6-1\choose3-1}={5\choose2}=10.$$

Give a ball to each chlid. Now you are left with the problem of how many ways to give 3 balls to 3 children. This is equal to the number of non-negative integer solutions of the equation $$x+y+z=3.$$ And the answer is $$\binom{3+3-1}{3}.$$

• To make your statement more precise, say something like "This is equal to the number of non-negative integer solutions of the equation $x + y + z = 3$." – N. F. Taussig Jan 26 '15 at 13:37
• Thanks, wasnt clear enough. – Ofir Schnabel Jan 26 '15 at 13:39