Question

If a and b are the greatest values of \( {}^{2n}C_{r} \) and \( {}^{(2n-1)}C_{r} \) respectively, then

• A. \( a=2b \)
• B. \( b=2a \)
• C. \( a=b \)
• D. \( a^{2}=2b^{2} \)

Hint:

Test with a simple value like \( n=2 \): \[ a = \max({}^4C_r) = {}^4C_2 = 6 \] \[ b = \max({}^3C_r) = {}^3C_1 = 3 \] Since \( 6 = 2(3) \), the relation \( a = 2b \) is verified instantly!

Answer 1

The Correct Option is A

Step 1: Recall where nCr is largest.
For $^mC_r$, the biggest value is at the middle. If $m$ is even, that is $r=\tfrac{m}{2}$; if $m$ is odd, it is at $r=\tfrac{m-1}{2}$.
Step 2: Find a.
For $^{2n}C_r$ the top $2n$ is even, so the greatest value is $a = {}^{2n}C_n$.
Step 3: Find b.
For $^{2n-1}C_r$ the top $2n-1$ is odd, so the greatest value is $b = {}^{2n-1}C_{n-1}$.
Step 4: Write a using factorials.
\[ a = \frac{(2n)!}{n!\,n!} = \frac{2n\,(2n-1)!}{n\,(n-1)!\,n!} \]
Step 5: Simplify the front.
The $\dfrac{2n}{n}$ becomes 2, and the rest is exactly $^{2n-1}C_{n-1}$. So $a = 2\cdot{}^{2n-1}C_{n-1}$.
Step 6: Conclude the relation.
Since $b = {}^{2n-1}C_{n-1}$, this gives \[ \boxed{a = 2b} \]