Question:medium
The sum \( S = \frac{1}{9!} + \frac{1}{3!7!} + \frac{1}{5!5!} + \frac{1}{7!3!} + \frac{1}{9!} \) is equal to:
The sum \( S = \frac{1}{9!} + \frac{1}{3!7!} + \frac{1}{5!5!} + \frac{1}{7!3!} + \frac{1}{9!} \) is equal to:
Show Hint
Whenever you see factorials in denominators summing to a constant (here \( 3+7=10, 5+5=10 \)), always try to multiply by that constant's factorial to convert the expression into a binomial sum.
Updated On: May 1, 2026
- \( \frac{2^{10}}{8!} \)
- \( \frac{2^9}{10!} \)
- \( \frac{2^7}{10!} \)
- \( \frac{2^6}{10!} \)
- \( \frac{2^5}{8!} \)
Show Solution
The Correct Option is B
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