Consider the sets A = {(x, y) in mathbb{R} ti | vedclass

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(C) Given sets are:$A: x^2 + y^2 = 25$$B: x^2 + 9y^2 = 144$$C: \{(x, y) \in \mathbb{Z} 	imes \mathbb{Z} : x^2 + y^2 \leq 4\}$To find $D = A \cap B$,w

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$17160$

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(C) Given sets are:$A: x^2 + y^2 = 25$$B: x^2 + 9y^2 = 144$$C: \{(x, y) \in \mathbb{Z} 	imes \mathbb{Z} : x^2 + y^2 \leq 4\}$To find $D = A \cap B$,we solve the equations of $A$ and $B$:$x^2 = 25 - y^2$Substitute into $B$: $(25 - y^2) + 9y^2 = 144$$8y^2 = 119 \Rightarrow y^2 = \frac{119}{8} \Rightarrow y = \pm \sqrt{\frac{119}{8}}$$x^2 = 25 - \frac{119}{8} = \frac{200 - 119}{8} = \frac{81}{8} \Rightarrow x = \pm \frac{9}{2\sqrt{2}}$Thus,$D$ contains $4$ points: $\left(\pm \frac{9}{2\sqrt{2}}, \pm \sqrt{\frac{119}{8}}ight)$. So,$|D| = 4$.Now,find elements of $C$ where $x, y \in \mathbb{Z}$ and $x^2 + y^2 \leq 4$:Possible integer pairs $(x, y)$ are:$(0, 0), (0, 1), (0, -1), (0, 2), (0, -2), (1, 0), (-1, 0), (2, 0), (-2, 0), (1, 1), (1, -1), (-1, 1), (-1, -1)$.Counting these,we get $|C| = 13$.The number of one-one functions from $D$ to $C$ is given by $P(n, r) = \frac{n!}{(n-r)!}$,where $n = |C| = 13$ and $r = |D| = 4$.Number of functions $= 13 	imes 12 	imes 11 	imes 10 = 17160$.

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