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I want to count the number of permutations of $4$ numbers, each in $\{1, 2, \ldots, 10\}$, that sum up to $10$. Can someone please show how to approach this problem? I want to know what is the best way to think about problems like this. Thank you.

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Do you mean four distinct numbers? Because in that case they have to be $1,2,3$, and $4$, and there are $4!=24$ permutations of that set of four. –  Brian M. Scott Mar 13 '13 at 15:23
    
They dont need to be distinct, e.g. (1 1 1 7), as long as they sum up to 10. –  cody Mar 13 '13 at 15:25

2 Answers 2

up vote 4 down vote accepted

This is a straightforward stars-and-bars problem: you want to count the solutions in positive integers to the equation $x_1+x_2+x_3+x_4=10$. There are

$$\binom{10-1}{4-1}=\binom93=84$$

such solutions. The reasoning behind the formula is explained pretty clearly in the linked article.

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If you're looking to eliminate the number of sums that differ only in their order of summands (e.g. $1+2+3+4=10$ and $4+1+2+3=10$) then this question is one on partitions, specifically the partition of $10$ into $4$ (non-zero) parts.

This input into Wolfram Alpha gives a nice visualization of the solution via Ferrers diagrams.

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