The number of elements in the set $\left\{x \in \mathbb{N}:^{20-2 x} c_{x-3} \in \mathbb{N}\right\}$ is
The number of elements in the set $\left\{x \in \mathbb{N}:^{20-2 x} c_{x-3} \in \mathbb{N}\right\}$ is
$3$
$4$
$5$
$6$
Solution
$\because^{20-2 x} C_{x-3} \in N$
So, $20-2 \mathrm{x}>0 \Rightarrow \mathrm{x} < 10$
and $x-3 \geq 0 \Rightarrow x \geq 3$
and $20-2 x \geq x-3 \Rightarrow x \leq \frac{23}{3}$
$\Rightarrow \mathrm{x} \leq 7 \quad\{\because \mathrm{x} \in \mathrm{N}\}$
Now, from above inequalities, we have
$3 \leq x \leq 7$
$\because \mathrm{x}$ is a natual number.
$\Rightarrow \mathrm{x}=3,4,5,6,7$
So, number of elements in the set $=5$