# Mathematics Combinatorics

If I have 10 BALLS and 3 boxes, how many possible number of solutions are there with a maximum number of 9 and minimum number of 1? numbers cannot be repeated.

Mathematically, we are finding the number of solutions of $x_1+x_2+x_3=10$ where $1\leq x_1,x_2,x_3\leq9$ and $x_1,x_2,x_3$ are integers.

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2,5,3? Those add up to ten and match your condition. Forgive me if I didnt understand your question right. –  Karim O. Jan 2 at 7:53
Never Mind. What we actually need is the number of possibilities that adds up to 10. –  suzy ong Jan 2 at 7:59
I guess [7,2,1] and [6,1,3] also work, not much combinations to work with –  Karim O. Jan 2 at 8:10

Straight enumeration is the way to go here, but you can simplify a little I think. Since you don't allow repeats, this is the same as $6$ times the number of ways to write $x_1 + x_2 + x_3 = 10$ with $x_i$ strictly increasing and greater than or equal to $1$. Subtracting $i$ from $x_i$, we see that's the same as counting the number of $x_1 + x_2 + x_3 = 4$ with the $x_i$ weakly increasing and greater than $0$. It's pretty quick to enumerate these, there are only $4$ of them, $(0,0,4),(0,1,3),(0,2,2),(1,1,2)$. So, there are $24$ ways to allocate the balls in your problem.

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Assume for the moment that we want $x_1<x_2<x_3$. The we necessarily have \eqalign{x_1&=1+y_1,\cr x_2&=(x_1+1)+y_2=2+y_1+y_2,\cr x_3&=(x_2+1)+y_3=3+y_1+y_2+y_3\cr} with $y_k\geq0$ $\>(1\leq k\leq 3)$. The condition $x_1+x_2+x_3=10$ then implies $$3y_1+2y_2+y_3=4\ ,\tag{1}$$ which automatically enforces $y_1+y_2+y_3\leq 6$, or $x_3\leq9$.

The nonnegative solutions of $(1)$ can be found by inspection: We obtain the triples $(0,0,4)$, $(0,1,2)$ and $(0,2,0)$ with $y_1=0$ and the single triple $(1,0,1)$ with $y_1=1$.

So there are $4$ solutions when the $x_k$ are in increasing order, or $24$ solutions when they can be in any order.

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