Given the sets:
\[ A = \{2, 3, 6, 7\}, \quad B = \{2, 5, 6, 8\} \]
The relation \( (a_1, b_1) \, R \, (a_2, b_2) \) is defined by:
\[ a_1 + a_2 = b_1 + b_2 \]
We list all possible valid pairs \((a_1, b_1)\) and \((a_2, b_2)\) satisfying the condition:
\[ \begin{aligned} 1. &(2, 4)R(6, 4) &\quad 2. &(2, 4)R(7, 5) \\ 3. &(2, 5)R(7, 4) &\quad 4. &(3, 4)R(6, 5) \\ 5. &(3, 5)R(6, 4) &\quad 6. &(3, 5)R(7, 5) \\ 7. &(3, 6)R(7, 4) &\quad 8. &(3, 4)R(7, 6) \\ 9. &(6, 5)R(7, 8) &\quad 10. &(6, 8)R(7, 5) \\ 11. &(7, 8)R(7, 6) &\quad 12. &(6, 8)R(6, 4) \\ 13. &(6, 6)R(6, 6) \end{aligned} \] × 2
Thus, the total number of such relations is:
\[ 24 + 1 = 25 \]
Step 1: Analyze the relation The sets are:
\[ A = \{2, 3, 6, 7\}, \quad B = \{4, 5, 6, 8\}. \]
The condition \((a_1, b_1) \, R \, (a_2, b_2)\) holds if:
\[ a_1 + a_2 = b_1 + b_2. \]
Step 2: Calculate valid pairs We evaluate all possible pairs \((a_1, b_1)\) and \((a_2, b_2)\) such that the condition holds.
Example pairs:
Total count: By systematically counting valid combinations, we find there are 24 such pairs. Additionally, there is one reflexive pair \((6, 6) \, R \, (6, 6)\).
Step 3: Total number of elements
Total number of elements in \(R = 24 + 1 = 25.\)
Final Answer: 25.
A relation R is defined in the set N as follows:
R = (x, y) : x = y - 3, y > 3
Then, which of the following is correct?
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:

In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by: