Consider the sets $$A = \{(x, y) \in \mathbb{R} \times \mathbb{R} : x^2 + y^2 = 25\}$$, $$B = \{(x, y) \in \mathbb{R} \times \mathbb{R} : x^2 + 9y^2 = 144\}$$, $$C = \{(x, y) \in \mathbb{Z} \times \mathbb{Z} : x^2 + y^2 \le 4\}$$, and $$D = A \cap B$$. The total number of one-one functions from the set D to the set C is:

The set $$A$$ is the circle with radius $$5$$ centred at the origin: $$x^2 + y^2 = 25$$.

The set $$B$$ is the ellipse $$x^2 + 9y^2 = 144$$ whose semi-axes are $$12$$ along the $$x$$-axis and $$4$$ along the $$y$$-axis.

To obtain $$D = A \cap B$$ we solve the two equations simultaneously.

From the circle, $$x^2 = 25 - y^2$$ $$-(1)$$. Substitute $$-(1)$$ into the ellipse:

$$25 - y^2 + 9y^2 = 144$$

$$25 + 8y^2 = 144$$

$$8y^2 = 119$$

$$y^2 = \frac{119}{8}$$, so $$y = \pm \sqrt{\frac{119}{8}}$$.

Using $$-(1)$$,

$$x^2 = 25 - \frac{119}{8} = \frac{81}{8}$$, hence $$x = \pm \sqrt{\frac{81}{8}} = \pm \frac{9\sqrt{2}}{4}$$.

Both $$x$$ and $$y$$ can take their signs independently, giving the four intersection points

$$\left(\pm\frac{9\sqrt{2}}{4}, \; \pm\sqrt{\frac{119}{8}}\right).$$

Therefore, $$|D| = 4$$.

Next, count the elements of $$C = \{(x, y) \in \mathbb{Z} \times \mathbb{Z} : x^2 + y^2 \le 4\}$$.

List all integer pairs inside or on the circle of radius $$2$$:

• For $$y = 0$$: $$x = -2, -1, 0, 1, 2$$ → $$5$$ points.
• For $$y = \pm 1$$: $$x^2 \le 3$$ → $$x = -1, 0, 1$$ → $$3 + 3 = 6$$ points.
• For $$y = \pm 2$$: $$x^2 \le 0$$ → $$x = 0$$ → $$1 + 1 = 2$$ points.

Total: $$5 + 6 + 2 = 13$$, so $$|C| = 13$$.

We need one-one (injective) functions from a $$4$$-element set $$D$$ to a $$13$$-element set $$C$$. For an injective function, choose the image of each element of $$D$$ as follows:

• First element: $$13$$ choices.
• Second element: $$12$$ choices.
• Third element: $$11$$ choices.
• Fourth element: $$10$$ choices.

Hence the total number of injective functions is the permutation

$$13 \times 12 \times 11 \times 10 = 17160.$$

Therefore, the required number of one-one functions is $$17160$$.

Option C is correct.