Question

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

• A. 15120
• B. 19320
• C. 17160
• D. 18290

Hint:

The number of one-one functions from a set with $m$ elements to a set with $n$ elements is given by $n \times (n-1) \times (n-2) \times \ldots \times (n-m+1)$.

Answer 1

The Correct Option is C

1. Define sets $A$ and $B$: - $A$: The circle defined by $x^2 + y^2 = 25$. - $B$: The ellipse defined by $\frac{x^2}{144} + \frac{y^2}{16} = 1$.
2. Determine the intersection $D = A \cap B$:
- Substitute $x^2 + y^2 = 25$ into the equation for $B$, which can be rewritten as $x^2 + 9y^2 = 144$: \[ x^2 + 9(25 - x^2) = 144 \] \[ x^2 + 225 - 9x^2 = 144 \] \[ -8x^2 = 144 - 225 \] \[ -8x^2 = -81 \] \[ x^2 = \frac{81}{8} \] \[ x = \pm \frac{9}{2\sqrt{2}} \] - Substitute the value of $x^2$ back into the equation for $A$ ($x^2 + y^2 = 25$): \[ y^2 = 25 - \frac{81}{8} \] \[ y^2 = \frac{200 - 81}{8} \] \[ y^2 = \frac{119}{8} \] \[ y = \pm \frac{\sqrt{119}}{2\sqrt{2}} \]
3. List the elements of set $D$: The intersection points are: \[ D = \left\{\left(\frac{9}{2\sqrt{2}}, \frac{\sqrt{119}}{2\sqrt{2}}\right), \left(\frac{9}{2\sqrt{2}}, -\frac{\sqrt{119}}{2\sqrt{2}}\right), \left(-\frac{9}{2\sqrt{2}}, \frac{\sqrt{119}}{2\sqrt{2}}\right), \left(-\frac{9}{2\sqrt{2}}, -\frac{\sqrt{119}}{2\sqrt{2}}\right)\right\} \] The cardinality of set $D$ is $|D| = 4$.
4. Identify set $C$: $C$ is the set of integer coordinate pairs $(x, y)$ such that $x^2 + y^2 \leq 4$. The elements are: \[ C = \{(0, 2), (2, 0), (0, -2), (-2, 0), (1, 1), (-1, -1), (1, -1), (-1, 1), (1, 0), (0, 1), (-1, 0), (0, -1), (0, 0)\} \] The cardinality of set $C$ is $|C| = 13$.
5. Calculate the number of one-to-one functions from set $D$ to set $C$: The number of injective functions from a set of size $k$ to a set of size $n$ is given by $P(n, k) = \frac{n!}{(n-k)!}$. In this case, $n=13$ and $k=4$. \[ \text{Number of one-to-one functions} = P(13, 4) = 13 \times 12 \times 11 \times 10 = 17160 \] The total number of one-to-one functions from set $D$ to set $C$ is 17160.