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We know that if $k,n$ are integers ($k\ge0$ and $n\ge1$), then the number of (integer nonnegative) solutions to the equation $a_1 + a_2 + ... + a_n = k$ is equal to $\binom{k + n - 1}{k}$ (Like placing k apples in n boxes).

What happens if we are just looking for solutions $(a_1,\cdots,a_n)$ such that $(a_1 > a_2 > a_3 > ...)$ ?

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    Then you're asking about the nummber of partitions of $k$ into $n$ distinct parts. There is a lot of literature about partitions on the web and in your favorite library. Short answer, though, is that results on partitions generally end up being much more complicated than just binomial coefficients. – 2017-02-15
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    So have you had a look for stuff about those partitions? – 2017-02-17
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    I'm voting to close this question as off-topic because OP has abandoned it. – 2017-02-19
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    I check some famous articles about partition of a number, but this problem is a bit different. In a common problem, k is expressed as a sum of exactly n distinct positive integers, without regard to order. What would happen if the first partition should be greater than the second one (and so on)? – 2017-02-20
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    I don't know what articles you looked at, but the number of partitions of $k$ into $n$ distinct parts is the number of solutions of $a_1+a_2+\cdots+a_n=k$ with $a_1>a_2>\cdots>a_n>0$, which seems to be pretty much exactly what you want. Type *partitions into distinct parts* into the internet, and report back to us on what you find. – 2017-02-20

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