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Given $(a, b, c, d)$ is a set of integers and $a \geq b \geq c \geq d \geq 0$. Find the number of solution sets for $a + b + c + d = 10$.

(This is problem from a competition, the answer key says its $23$, wherein my answer is $455$)

Solution

$x_1 + x_2 +1 + x_3 + 2 + x_4 + 3 = 10$

$x_1 + x_2 + x_3 + x_4 = 16$

$_{15}C_3$ = 455

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    $\begingroup$ Edit the question to show your solution so we can critique it. $\endgroup$ Feb 20, 2018 at 16:44
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    $\begingroup$ Your solution is incomrehensible. What are the $x_1x_2,x_3,x_4$ and why do when you add 1,2 and 3 to the sum do you get $1$. Why is the number of solutions to $x_1 + x_2 + x_3 + x_4 = 16$ $_{15}C_3$? $\endgroup$
    – fleablood
    Feb 20, 2018 at 16:53
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    $\begingroup$ You should explain what you are doing instead of writing down lines of equations with no comments at all. $\endgroup$
    – Christoph
    Feb 20, 2018 at 17:00
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    $\begingroup$ The answer is the number of partitions of 10 with 4 or fewer parts, which is 23. $\endgroup$
    – Jens
    Feb 20, 2018 at 17:08
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    $\begingroup$ @Kusavil: Not that I know of. I simply have an algorithm to find these partitions. It's based on info I found here. $\endgroup$
    – Jens
    Mar 2, 2018 at 17:07

2 Answers 2

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The solutions are: $$\begin{align}(d,c,b,a)= &(0,0,0,10) \cdots (0,0,5,5) \Rightarrow 6 \\ &(0,1,1,8) \cdots (0,1,4,5) \Rightarrow 4 \\ &(0,2,2,6), (0,2,4,4) \Rightarrow 3 \\ &(0,3,3,4) \Rightarrow 1 \\ &(1,1,1,7) \cdots (1,1,4,4) \Rightarrow 4 \\ &(1,2,2,5), (1,2,3,4) \Rightarrow 2 \\ &(1,3,3,3) \Rightarrow 1 \\ &(2,2,2,4),(2,2,3,3) \Rightarrow 2 \end{align}$$ Hence, there are $23$ solutions.

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    $\begingroup$ is there some easy way or formula to do this? What, if instead of $a,b,c,d$ we had $x_1,...,x_k$ and we were to solve $x_1+\dots+x_k = n$, where $0 \le x_1 \le \dots \le x_k$? $\endgroup$
    – Kusavil
    Feb 20, 2018 at 17:36
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$a \geq b \geq c \geq d \geq 0$ give easily as follows $$\begin{cases}a=10\text{ gives } 1\\a=9\text{ gives } 1\\a=8\text{ gives }2\\a=7\text{ gives }3\\a=6\text{ gives }4\\a=5\text{ gives }5\\a=4\text{ gives }5\\a=3\text{ gives }2 \end{cases}$$ examples: $8200,8110\text{ are the two solutions for a=8 }\\ 5500,5410,5320,5311,5221\text{ are the five solutions for a=5 }$

Thus there are $23$ solutions.

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