First slide

Combinations

Question

If $a$ and $b$ are the greatest values of $^{2 n} C_{r} {and}^{2 n - 1} C_{r}$ respectively. Then

Moderate

A

$a = 2 b$
B

$b = 2 a$
C

$a = b$
D

none of these

Solution

We know that the greatest value of $^{n} C_{r}$ is given by
$\{\begin{cases} ^{n} C_{n / 2} if n is even \\ \frac{^{n} C_{n - 1}}{2} or \frac{^{n} C_{n + 1}}{2}, if n is odd \end{cases}$
$∴ a =$ greatest value of $^{2 n} C_{r} =^{2 n} C_{n}$
and,
$b =$ Greatest value of $^{2 n - 1} C_{r} =^{2 n - 1} C_{n - 1}$
$\Rightarrow \frac{a}{b} = \frac{^{2 n} C_{n}}{^{2 n - 1} C_{n - 1}} = \frac{{\frac{2 n}{n}}^{2 n - 1} C_{n - 1}}{2 n - 1_{C_{n - 1}}} = 2 \Rightarrow a = 2 b$

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