Find the product of all odd natural numbers less than 5000.

### Concept & Formula This problem requires **factorial manipulation**. The factorial of a number ($n!$) represents the product of all positive integers up to $n$. To isolate the product of only the odd numbers, you must take the full factorial and divide out the product of all the even numbers. $$\text{Product of Odds} = \frac{\text{Total Product } (n!)}{\text{Product of Evens}}$$ ### Step-by-Step Solution * **Given:** We need the product of odds up to 4999. Let $P = 1 \times 3 \times 5 \times ... \times 4999$. * **Calculation:** Start with the total factorial equation for 5000: $$5000! = (1 \times 3 \times 5 \times ... \times 4999) \times (2 \times 4 \times 6 \times ... \times 5000)$$ * Substitute our target variable $P$ into the equation: $$5000! = P \times (2 \times 4 \times 6 \times ... \times 5000)$$ * Now, look specifically at the evens product: $(2 \times 4 \times 6 \times ... \times 5000)$. * Because every number in that series is even, we can factor out a 2 from every single term. There are exactly 2500 terms in that series. $$\text{Evens} = (2 \times 1) \times (2 \times 2) \times (2 \times 3) \times ... \times (2 \times 2500)$$ $$\text{Evens} = 2^{2500} \times (1 \times 2 \times 3 \times ... \times 2500)$$ * The remaining series $(1 \times 2 \times ... \times 2500)$ is simply $2500!$. So we can rewrite the evens product as: $$\text{Evens} = 2^{2500} \times 2500!$$ * Substitute this simplified evens expression back into the main equation: $$5000! = P \times (2^{2500} \times 2500!)$$ * Finally, isolate $P$ (the product of odds) by dividing: $$P = \frac{5000!}{2^{2500} \times 2500!}$$ ### Exam Strategy & Shortcut Memorize this derived standard formula: For any even target number $2n$, the product of all odd numbers less than $2n$ is always equal to **$\frac{(2n)!}{2^n \times n!}$**. In this question, $2n = 5000$, which makes $n = 2500$. Plugging $n=2500$ directly into the formula gives $\frac{5000!}{2^{2500} \times 2500!}$ instantly, bypassing all algebraic derivation during the exam. ### Common Pitfall When students try to factor out the $2$ from the sequence of evens $(2 \times 4 \times 6 \times ...)$, they often mistakenly pull it out as a single constant factor, arriving at $2 \times (1 \times 2 \times 3 ...)$ which equals $2 \times 2500!$. They forget that because the numbers are being multiplied, a $2$ must be extracted from *each individual term*, resulting in $2^{2500}$. ### Final Answer Therefore, the correct answer is **$\frac{5000!}{2^{2500} \times 2500!}$**.