Question

\( \frac{1}{9!} + \frac{1}{3!7!} + \frac{1}{5!5!} + \frac{1}{7!3!} + \frac{1}{9!} \) is equal to

• A. \( \frac{2^9}{10!} \)
• B. \( \frac{2^{10}}{8!} \)
• C. \( \frac{2^{11}}{9!} \)
• D. \( \frac{2^{10}}{7!} \)
• E. \( \frac{2^8}{9!} \)

Hint:

Symmetric factorial expressions often come from binomial expansions.

Answer 1

The Correct Option is A

Concept: Recognize binomial expansion identity.

Step 1: Observe symmetry of terms.
\[ \frac{1}{0!9!}, \frac{1}{1!8!}, \frac{1}{2!7!}, \frac{1}{3!6!}, \frac{1}{4!5!} \] This resembles binomial coefficients.

Step 2: Recall identity.
\[ \sum \frac{1}{r!(n-r)!} = \frac{2^n}{n!} \]

Step 3: Here \( n=9 \).
\[ \sum_{r=0}^{9} \frac{1}{r!(9-r)!} = \frac{2^9}{9!} \]

Step 4: Given expression is half of full expansion.
So: \[ = \frac{1}{2} \cdot \frac{2^9}{9!} \]

Step 5: Simplify.
\[ = \frac{2^8}{9!} = \frac{2^9}{10!} \]