Key Concepts and Formulas

• Intersection of Sets: The intersection of two sets $A$ and $B$, denoted by $A \cap B$, is the set of all elements that are in both $A$ and $B$.
• Geometric Representation of Sets:
  • $x^2+y^2=r^2$ represents a circle centered at the origin with radius $r$.
  • $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ represents an ellipse centered at the origin with semi-major/minor axes $a$ and $b$.
  • $x^2+y^2 \leq r^2$ represents the region inside and on the circle of radius $r$ centered at the origin.
• Number of One-One Functions: If $X$ is a set with $m$ elements and $Y$ is a set with $n$ elements, the number of one-one functions from $X$ to $Y$ is given by $P(n, m) = \frac{n!}{(n-m)!}$, provided $n \geq m$. If $n < m$, the number of one-one functions is 0.

Step-by-Step Solution

Step 1: Determine the cardinality of set $D$

The set $D$ is the intersection of sets $A$ and $B$, i.e., $D = A \cap B$. The equations defining $A$ and $B$ are: $A=\left\{(x, y) \in \mathbb{R} \times \mathbb{R}: x^2+y^2=25\right\}$ $B=\left\{(x, y) \in \mathbb{R} \times \mathbb{R}: x^2+9 y^2=144\right\}$

Set $A$ is a circle with radius 5 centered at the origin. Set $B$ is an ellipse. Dividing the equation by 144, we get $\frac{x^2}{144} + \frac{y^2}{16} = 1$, which is an ellipse with semi-major axis $a=12$ along the x-axis and semi-minor axis $b=4$ along the y-axis.

To find the points in $D$, we need to solve the system of equations:

1. $x^2+y^2=25$
2. $x^2+9y^2=144$

Subtract equation (1) from equation (2): $(x^2+9y^2) - (x^2+y^2) = 144 - 25$ $8y^2 = 119$ $y^2 = \frac{119}{8}$

Substitute $y^2$ back into equation (1): $x^2 + \frac{119}{8} = 25$ $x^2 = 25 - \frac{119}{8} = \frac{200 - 119}{8} = \frac{81}{8}$

Now we find the possible values for $x$ and $y$: $x = \pm \sqrt{\frac{81}{8}} = \pm \frac{9}{2\sqrt{2}} = \pm \frac{9\sqrt{2}}{4}$ $y = \pm \sqrt{\frac{119}{8}} = \pm \frac{\sqrt{119}}{2\sqrt{2}} = \pm \frac{\sqrt{238}}{4}$

Since $x^2$ and $y^2$ are positive, there are real solutions for $x$ and $y$. The $\pm$ signs indicate that for each value of $x$, there are two possible values of $y$, and vice versa. The four points of intersection are: $\left(\frac{9\sqrt{2}}{4}, \frac{\sqrt{238}}{4}\right), \left(\frac{9\sqrt{2}}{4}, -\frac{\sqrt{238}}{4}\right), \left(-\frac{9\sqrt{2}}{4}, \frac{\sqrt{238}}{4}\right), \left(-\frac{9\sqrt{2}}{4}, -\frac{\sqrt{238}}{4}\right)$. These are four distinct real points. Therefore, the cardinality of set $D$ is $|D|=4$.

Step 2: Determine the cardinality of set $C$

The set $C$ is defined as: $C=\left\{(x, y) \in \mathbb{Z} \times \mathbb{Z}: x^2+y^2 \leq 4\right\}$ This means we need to find all pairs of integers $(x,y)$ that lie inside or on the circle $x^2+y^2=4$ (a circle of radius 2 centered at the origin).

We list the integer points systematically:

• If $x=0$: $y^2 \leq 4 \implies y \in \{-2, -1, 0, 1, 2\}$. This gives points: $(0,-2), (0,-1), (0,0), (0,1), (0,2)$ (5 points).
• If $x=1$: $1+y^2 \leq 4 \implies y^2 \leq 3 \implies y \in \{-1, 0, 1\}$. This gives points: $(1,-1), (1,0), (1,1)$ (3 points).
• If $x=-1$: $1+y^2 \leq 4 \implies y^2 \leq 3 \implies y \in \{-1, 0, 1\}$. This gives points: $(-1,-1), (-1,0), (-1,1)$ (3 points).
• If $x=2$: $4+y^2 \leq 4 \implies y^2 \leq 0 \implies y=0$. This gives point: $(2,0)$ (1 point).
• If $x=-2$: $4+y^2 \leq 4 \implies y^2 \leq 0 \implies y=0$. This gives point: $(-2,0)$ (1 point).
• If $|x| > 2$, then $x^2 > 4$, so $x^2+y^2 > 4$ for any real $y$, thus no integer points exist.

The total number of integer points in set $C$ is $5 + 3 + 3 + 1 + 1 = 13$. Therefore, the cardinality of set $C$ is $|C|=13$.

Step 3: Calculate the total number of one-one functions

We need to find the number of one-one functions from set $D$ to set $C$. The domain is set $D$ with $|D|=m=4$. The codomain is set $C$ with $|C|=n=13$.

Since $n \geq m$ ($13 \geq 4$), the number of one-one functions is given by the permutation formula $P(n, m)$: Number of one-one functions $= P(13, 4) = \frac{13!}{(13-4)!} = \frac{13!}{9!}$ $P(13, 4) = 13 \times 12 \times 11 \times 10$ $P(13, 4) = 156 \times 110$ $P(13, 4) = 17160$

Common Mistakes & Tips

• Domain of Coordinates: Be careful to distinguish between $\mathbb{R} \times \mathbb{R}$ (real numbers) and $\mathbb{Z} \times \mathbb{Z}$ (integers). This distinction is critical for determining the elements of sets $D$ and $C$.
• Integer Points: When dealing with sets defined by inequalities and integer coordinates (like set $C$), systematically list all possible integer values to avoid missing any points.
• Permutation Formula: Ensure you are using the correct formula for one-one functions, which is $P(n, m)$, not combinations or other permutation formulas. Remember that the order of mapping matters for functions.

Summary

The problem requires finding the number of one-one functions from set $D$ to set $C$. First, we determined that set $D$, the intersection of a circle and an ellipse, contains 4 distinct real points. Next, we identified the integer coordinate points within or on a circle of radius 2, finding that set $C$ contains 13 such points. Finally, using the formula for the number of one-one functions, $P(n, m)$, with $n=|C|=13$ and $m=|D|=4$, we calculated the total number of one-one functions to be $P(13, 4) = 17160$.

The final answer is \boxed{17160}.