[Math] The sum of three positive integers is $20$. Find the probability that they form the sides of a triangle.

permutationsprobability

The sum of three positive integers is $20$. Find the probability that they form the sides of a triangle.

Let $a,b,c$ be there positive integers. So, $a+b+c=20$

Total number of solutions would be $^{19}C_2=171$

For $a,b,c$ to form a triangle $0\lt a,b,c\le9$

So, number of such solutions= coefficient of $x^{20}$ in $(x+x^2+x^3+…+x^9)^3=$ coefficient of $x^{17}$ in $(1+x+x^2+…+x^8)^3=$ coefficient of $x^{17}$ in $(1-x^9)^3(1-x)^{-3}=1\times^{19}C_2-3\times^{10}C_2=36$

So, the required probability is $\frac{36}{171}$ but the answer given is $\frac8{33}$

In the hint, they have written total combinations of $a,b,c=\frac{144}{6}+\frac{27}{3}=33$, and favorable combinations of $a,b,c=\frac{12}{3}+\frac{24}{6}=8$

I think they have split $171$ as $144+27$ and $36$ as $24+12$ but why? and why to divide them with $6$ or $3$?

Best Answer

It is a bad question as they have not described how the sides have been chosen, but their answer almost reveals what they intended:

  • There are $171$ compositions of $20$ into three positive integer parts (any order). Of these,
    • $144$ compositions have the three parts distinct
    • $27$ compositions have two of the parts equal and the third distinct
    • $0$ compositions have all three parts equal ($20$ is not divisible by $3$)
  • So removing duplicates which are the same but in a different order, there are
    • $\frac{144}{3!}=24$ partitions which have the three parts distinct
    • $\frac{27}{3}=9$ partitions which have two of the parts equal and the third distinct
    • $0$ partitions have all three parts equal ($20$ is not divisible by $3$)
    • making $24+9+0=33$ for the number of partitions of $20$ into three positive integer parts

The $36 \to 8$ is the same but where the two smaller numbers add up to more than the largest number

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