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Let $\mathbb{N}$ be the set of positive integers. The number of distinct triplets $(x, y, z)$ satisfying $x, y, z \in \mathbb{N}, x < y < z$ and $x+y+z=12$ is
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The correct answer is:
7
Given $\mathbf{N}$ is the set of positive integers with triplets $(x, y, z)$.
The combination of $x, y, z$ should satisfy $x < y < z$ and $x+y+z=12$.
Required combinations are shown below.
$\begin{aligned}
& 1 < 2 < 9,2 < 3 < 7,3 < 4 < 5,1 < 5 < 6,2 < 4 < 6, \\
& 1 < 3 < 8,1 < 4 < 7
\end{aligned}$
So, there are total 7 combinations.
The combination of $x, y, z$ should satisfy $x < y < z$ and $x+y+z=12$.
Required combinations are shown below.
$\begin{aligned}
& 1 < 2 < 9,2 < 3 < 7,3 < 4 < 5,1 < 5 < 6,2 < 4 < 6, \\
& 1 < 3 < 8,1 < 4 < 7
\end{aligned}$
So, there are total 7 combinations.
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