First slide

Combinations

Question

$^{2 n} C_{r} (0 \leq r \leq 2 n)$ is greatest when r is equal to

Moderate

A

$\frac{n}{2}$
B

$\frac{n + 1}{2}$
C

r = n
D

None of these

Solution

We have, $\frac{^{2 n} C_{r}}{^{2 n} C_{r - 1}} = \frac{2 n - r + 1}{r}$ (1)
For $^{2 n} C_{r}$ to be greatest $\frac{2 n - r + 1}{r} \geq 1$
$\Rightarrow 2 n - r + 1 \geq r \Rightarrow 2 r \leq 2 n + 1$
$\Rightarrow r \leq n + \frac{1}{2}$ (2)
From (1), $\frac{^{2 n} C_{r + 1}}{^{2 n} C_{r}} = \frac{2 n - (r + 1) + 1}{r + 1} = \frac{2 n - r}{r + 1}$ .
For $^{2 n} C_{r}$ to be greatest $\frac{2 n - r}{r + 1} \leq 1$
$\Rightarrow 2 n - r \leq r + 1 \Rightarrow 2 r \geq 2 n - 1$
$\Rightarrow r \geq n - \frac{1}{2}$ (3)
From (2) and (3), we get $n - \frac{1}{2} \leq r \leq n + \frac{1}{2}$
⇒ r = n (since r is a positive integer)
Hence, $^{2 n} C_{r}$ is greatest when r = n.

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