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Question #41030

Difficulty: 8/10 | 🟢 Free

The Question

Let $N$ be the exponent of the highest power of $3$ dividing $100!$ , and $k$ be the exponent of the highest power of $5$ dividing $100!$ . A subset $A$ is chosen at random from the power set of $S = \{1, 2, \dots, N\}$ . The probability that the number of elements in $A$ is strictly greater than $k$ is $p$ . Then $p$ is equal to:

Options

A.
$\frac{1}{2} - \frac{\binom{48}{24}}{2^{49}}$
B.
$\frac{1}{2} + \frac{\binom{48}{24}}{2^{49}}$
C.
$\frac{1}{2} - \frac{\binom{48}{24}}{2^{48}}$
D.
$\frac{1}{2} + \frac{\binom{48}{24}}{2^{48}}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Complementary Counting|Combinatorial Counting (P&C Integration)|Exponent of Prime in n!|Greatest Binomial Coefficient

Why This Question Is Tricky

Common Mistakes

The problem fuses Legendre's Formula with combinatorial probability, creating a computational bottleneck if approached directly.

The core trap lies in the boundary condition of the summation: students must recognize that $k = N/2$ , making $\binom{N}{k}$ the greatest binomial coefficient.

When applying complementary counting to evaluate $\sum_{r=25}^{48} \binom{48}{r}$ , a common mistake is failing to isolate and halve the middle term $\binom{48}{24}$ , leading to incorrect probability offsets (Options B and C).

This tests both parity symmetry in binomial expansions and precision in boundary cases.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 11
Concept	Complementary Counting
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free