Question

The product of first \( n \) odd natural numbers is :

• A. \( (^{2n}C_n) (^nP_n) \)
• B. \( (1/2) (^{2n}C_n) (^nP_n) \)
• C. \( (1/2^n) (^{2n}C_n) (^nP_n) \)
• D. \( (1/2^{2n}) (^{2n}C_n) (^nP_n) \) Correct Answer: (C) \( (1/2^n) (^{2n}C_n) (^nP_n) \) Solution:

Hint:

To verify such general formulas quickly in an exam, test with \( n=2 \).
Product of first 2 odds: \( 1 \cdot 3 = 3 \).
Using option (C) for \( n=2 \):
\( \frac{1}{2^2} \cdot ^4C_2 \cdot ^2P_2 = \frac{1}{4} \cdot 6 \cdot 2 = \frac{12}{4} = 3 \).
Since the values match, the formula is correct. This is much faster than deriving factorials.

Answer 1

The Correct Option is C

Step 1: Write the product of odd numbers.
We want $P = 1\cdot 3\cdot 5\cdots(2n-1)$.

Step 2: Bring in the even numbers.
Multiply top and bottom by all the even numbers up to $2n$. The top becomes every integer up to $2n$, which is $(2n)!$.
\[ P = \frac{(2n)!}{2\cdot 4\cdot 6\cdots(2n)} \]

Step 3: Simplify the even product.
Each even number is $2$ times something, and there are $n$ of them, so the bottom is $2^n\, n!$.
\[ P = \frac{(2n)!}{2^n\, n!} \]

Step 4: Match with the given symbols.
Note $\,^{2n}C_n \cdot \,^nP_n = \dfrac{(2n)!}{n!\,n!}\cdot n! = \dfrac{(2n)!}{n!}$. So $P$ is just this divided by $2^n$.
\[ P = \frac{1}{2^n}\left(\,^{2n}C_n\right)\left(\,^nP_n\right) \]
That is option 3.
\[ \boxed{\dfrac{1}{2^n}\left(\,^{2n}C_n\right)\left(\,^nP_n\right)} \]