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Question #41024

Difficulty: 8/10 | 🟢 Free

The Question

A light ray travels along the line $L: \frac{x}{3} = \frac{y}{4} = \frac{z}{12}$ and intersects a series of parallel glass plates. The surfaces of the plates lie on the planes $\Pi_k: 3x + 4y = 25k$ for $k = 1, 2, \dots, 40$ . Let the intersection points of the ray with these planes be $P_1, P_2, \dots, P_{40}$ respectively. A subset of 7 points is to be chosen from these 40 points such that the spatial distance between the orthogonal projections of any two chosen points on the plane $\Pi: 3x + 4y = 0$ is at least 39. Find the number of ways to choose these 7 points.

Options

A.
$\binom{22}{7}$
B.
$\binom{28}{7}$
C.
$\binom{34}{7}$
D.
$\binom{24}{7}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Combinatorial Counting|Gap Method|Distance Formula in 3D|Angle between line and plane

Why This Question Is Tricky

Common Mistakes

This problem fuses 3D Coordinate Geometry with Combinatorics.

The core trap lies in the spatial orientation of the points: students often calculate the distance along the line of intersection ( $d=13$ ) and apply the distance constraint directly, yielding an incorrect index gap of $|i-j| \ge 3$ (leading to the trap answer $\binom{28}{7}$ ).

By forcing the constraint onto the orthogonal projections, the effective distance between points changes ( $d'=12$ ), which alters the combinatorial threshold to $|i-j| \ge 4$ .

This tests whether the student can successfully translate a 3D geometric projection into a 1D discrete combinatorial sequence before applying the Gap Method.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 11
Concept	Combinatorial Counting (P&C Integration)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free