How many set of positive integers $\{a,b,c,d\}$ are there such that $ a\leq b\leq c\leq d\leq 50$ and $a+b+c+d=100$?
I was thinking about using stars and bars, and it seems to work if there were only three variables: If $a\leq b\leq c\leq 50$ and $a+b+c=100$, then we can define $x=50-a,y=50-b,z=50-c$ such that $x,y,z\leq 50$ and $x+y+z=3\times 50-(a+b+c)=50$. Then we can simply use stars and bars to find the number of triples of $\{x,y,z\}$, each of which corresponds to a unique $\{a,b,c\}$.
If $x\neq y,y\neq z$, and $z\neq x$, then only one out of the six sets consisting of $x,y,$ and $z$ is listed from the smallest to the largest. Since $50$ is not a multiple of $3$, $x=y=z$ is not possible. If two of the three elements are the same, then this element can be any integer between $1$ and $24$, inclusive. Since it can be either $x=y\neq z$, $x=z\neq y$, or $y=z\neq x$, each case appears three times. Thus, the total number of ordered triples $\{a,b,c\}$ is $\dfrac{C_{49}^2-24\times 3}{6}+24$.
However, when there's a fourth variable $d$, this method doesn't seem to work. Are there any other ways to circumvent it? Sorry for my poor English.
from Hot Weekly Questions - Mathematics Stack Exchange
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