2024 CSAT — Q49

The universe is $\{1,2,3,4,5,6,7\}$ and the chain is $x > 2y > 3z$. Anchor on the smallest, $z$, since $3z$ grows fastest.

$z = 1$: need $2y > 3 \Rightarrow y \ge 2$, and $x > 2y$ with $x \le 7$.

• $y = 2$: $2y = 4$, so $x \in \{5,6,7\}$ — 3 triplets.
• $y = 3$: $2y = 6$, so $x = 7$ — 1 triplet.
• $y \ge 4$: $2y \ge 8 > 7$, no $x$.

$z = 2$: need $2y > 6 \Rightarrow y \ge 4$, so $2y \ge 8 > 7$ — no valid $x$. 0.

$z \ge 3$: need $2y > 9 \Rightarrow y \ge 5$, so $x > 2y \ge 10$ — impossible. 0.

Total $= 3 + 1 = 4$.

Answer: (d) Four triplets.

Chained strict inequalities over a tiny set are best handled by anchoring on the variable with the largest coefficient ($z$, weighted $3$) and working outward, because that variable is pinned almost immediately ($z=1$ is forced — $z=2$ already needs $2y>6$, leaving no room for $x\le7$). Then for each $(z,y)$ simply count the admissible $x$. The two traps are stopping after the obvious extreme triplet, and forgetting that $y$ can take more than one value when $z=1$. Bound first, then enumerate the small residual.