Consider the sets $A = \{(x, y) \in R \times R : x^2 + y^2 = 25\}$, $B = \{(x, y) \in R \times R: x^2 + 9y^2 = 144\}$, $C = \{(x, y) \in Z \times Z: x^2 + y^2 \leq 4\}$ and $D = A \cap B$. The total number of one-one functions from the set $D$ to the set $C$ is:
Correct Answer:
C
Explanation
$A = \{(x, y) \in R \times R : x^2 + y^2 = 25\}$, $B = \{(x, y) \in R \times R: x^2 + 9y^2 = 144\}$
$x^2 + 9y^2 - (x^2 + y^2) = 144 - 25$
Plug in $y^2 = \frac{119}{8}$ into either equation to find $x$.
$x^2 = 25 - \frac{119}{8}$
$x^2 = \frac{200 - 119}{8}$
$x^2 = \frac{81}{8}$
$x = \pm \sqrt{\frac{81}{8}}$, $y = \pm \sqrt{\frac{119}{8}}$
Now, $C = \{(x, y) \in Z \times Z: x^2 + y^2 \leq 4\}$
Valid points are $(-2, 0), (-1, -1), (-1, 0), (-1, 1), (0, -2), (0, -1), (0, 0), (0, 1), (0, 2), (1, -1), (1, 0), (1, 1)$
$\therefore$ Total valid points in $C = 13$
$\Rightarrow$ There are 4 distinct real points in set $D$
$\therefore$ The number of one-one functions from $D$ to $C$
$\Rightarrow {}^{13}P_4 \Rightarrow \frac{13!}{(13-4)!} = \frac{13!}{9!} = 17160$