Step 1: Understand relation
Set $A = {2,3,4,5,6}$
Relation on $A \times A$:
$(x,y),R,(z,w)$ if:
Total elements = number of quadruples $(x,y,z,w)$ satisfying both conditions
Step 2: Count pairs for $x \mid z$
Check each $x$:
Total valid $(x,z)$ pairs:
$ 3 + 2 + 1 + 1 + 1 = 8 $
Step 3: Count pairs for $y \le w$
For each $y$, count $w \ge y$:
Total valid $(y,w)$ pairs:
$ 5 + 4 + 3 + 2 + 1 = 15 $
Step 4: Total elements in relation
Since independent:
$ \text{Total} = 8 \times 15 = 120 $
Final Answer:
$\boxed{120}$
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.
If $ \left(\frac{1}{^{15}C_0} + \frac{1}{^{15}C_1}\right)$ $\left(\frac{1}{^{15}C_1} + \frac{1}{^{15}C_2}\right)$ $\dots$ $ \left(\frac{1}{^{15}C_{12}} + \frac{1}{^{15}C_{13}}\right)$ = $\frac{\alpha^{13}}{^{14}C_0 ^{14}C_1 \dots ^{14}C_{12}} $, then $30\alpha$ is equal to __________.
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.