Consider the sets A = (x,y)RR : x^2+y^2=25, B = (x,y)RR : x^2+9y^2=144, C = (x,y)ZZ : x^2+…

Question

Consider the sets $A = \{(x,y)\in\mathbb{R}	imes\mathbb{R} : x^2+y^2=25\}$, $B = \{(x,y)\in\mathbb{R}	imes\mathbb{R} : x^2+9y^2=144\}$, $C = \{(x,y)\in\mathbb{Z}	imes\mathbb{Z} : x^2+y^2\leq4\}$, and $D=A\cap B$. The total number of one-one functions from the set $D$ to the set $C$ is:

Accepted Answer

Find $D=A\cap B$: Subtracting $x^2+y^2=25$ from $x^2+9y^2=144$: $8y^2=119 \Rightarrow y^2=\dfrac{119}{8}$, and $x^2=25-\dfrac{119}{8}=\dfrac{81}{8}$. Both give two real values each ($\pm$), so there are $2\times2=4$ real points in $D$. Thus $n(D)=4$.

Find $C$: integer lattice points with $x^2+y^2\leq4$:
$$(0,0),(\pm1,0),(0,\pm1),(\pm2,0),(0,\pm2),(\pm1,\pm1)$$
Count: $1+2+2+2+2+4 = 13$, so $n(C)=13$.

One-one functions from $D$ (4 elements) to $C$ (13 elements):
$$^{13}P_4 = \frac{13!}{9!} = 13\times12\times11\times10 = 17160$$

Solutions · 1