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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?

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    $\begingroup$ Are the balls the same or different? $\endgroup$ – Ofir Schnabel Jan 26 '15 at 10:29
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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?

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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 $$ {6-1\choose3-1}={5\choose2}=10. $$

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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 $$\binom{3+3-1}{3}.$$

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

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