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: Understanding the Concept:
The product of the first \( n \) odd numbers is expressed as \( 1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n-1) \). In mathematics, this is sometimes called the "double factorial" of \( (2n-1) \), written as \( (2n-1)!! \).
Since factorials represent products of consecutive integers, we can represent a product of only odd integers by taking a full factorial of all integers up to \( 2n \) and then dividing out the even integers. This approach allows us to use standard combinatorial notations like \( ^nC_r \) (combinations) and \( ^nP_n \) (permutations, which is simply \( n! \)) to represent the result.

Step 2: Key Formula or Approach:
The core algebraic trick is:
\[ \text{Product of odds} = \frac{\text{Product of all integers up to } 2n}{\text{Product of even integers up to } 2n} \]
We know:
1. Product of all integers up to \( 2n = (2n)! \)
2. Product of even integers = \( 2 \cdot 4 \cdot 6 \cdot \dots \cdot (2n) = 2^n \cdot (1 \cdot 2 \cdot 3 \cdot \dots \cdot n) = 2^n \cdot n! \)
3. \( ^{2n}C_n = \frac{(2n)!}{n! \cdot n!} \)
4. \( ^nP_n = n! \)

Step 3: Detailed Explanation:
Step 3.1: Constructing the odd product.
Let \( P = 1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n-1) \).
Multiply and divide by \( 2 \cdot 4 \cdot 6 \cdot \dots \cdot (2n) \):
\[ P = \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot \dots \cdot (2n-1) \cdot 2n}{2 \cdot 4 \cdot 6 \cdot \dots \cdot 2n} \]
The numerator is \( (2n)! \).
The denominator is \( 2^n(n!) \).
So, \( P = \frac{(2n)!}{2^n n!} \).
Step 3.2: Converting to the desired notation.
The options involve \( ^{2n}C_n \). Let's expand that:
\[ ^{2n}C_n = \frac{(2n)!}{n! n!} \]
If we multiply this by \( ^nP_n \) (which is \( n! \)):
\[ (^{2n}C_n) \cdot (^nP_n) = \frac{(2n)!}{n! n!} \cdot n! = \frac{(2n)!}{n!} \]
Step 3.3: Comparing the two results.
Our target product is \( P = \frac{(2n)!}{2^n n!} \).
Our notation product is \( K = \frac{(2n)!}{n!} \).
Clearly, \( P = \frac{K}{2^n} \).
Substituting \( K \) back:
\[ P = \frac{1}{2^n} \left( ^{2n}C_n \cdot ^nP_n \right) \]
This matches Option (C).

Step 4: Final Answer:
By expressing the product of odd integers as a ratio of full factorials and then substituting the definitions of permutations and combinations, we find the relation \( \frac{1}{2^n} (^{2n}C_n) (^nP_n) \). This confirms Option (C) is correct.