Question:medium
\( \frac{1}{9!} + \frac{1}{3!7!} + \frac{1}{5!5!} + \frac{1}{7!3!} + \frac{1}{9!} \) is equal to
\( \frac{1}{9!} + \frac{1}{3!7!} + \frac{1}{5!5!} + \frac{1}{7!3!} + \frac{1}{9!} \) is equal to
Show Hint
Symmetric factorial expressions often come from binomial expansions.
Updated On: May 1, 2026
- \( \frac{2^9}{10!} \)
- \( \frac{2^{10}}{8!} \)
- \( \frac{2^{11}}{9!} \)
- \( \frac{2^{10}}{7!} \)
- \( \frac{2^8}{9!} \)
Show Solution
The Correct Option is A
Solution and Explanation
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