Step 1: Understanding the Concept:
We need to count pairs \( (a, b) \) satisfying several conditions.
1. \( a \in A \) (multiples of 3 in \( [1, 700] \)).
2. \( b \in B \) (multiples of 7 in \( [1, 300] \)).
3. \( a + b \) is even, which means \( a \) and \( b \) must have the same parity (both even or both odd).
4. \( a \neq b \).
Step 2: Key Formula or Approach:
1. List the count of odd and even elements in A and B.
2. Calculate pairs (Odd, Odd) and (Even, Even).
3. Subtract cases where \( a = b \).
Step 3: Detailed Explanation:
Set A: Multiples of 3 between 1 and 700.
Max element: \( 699 = 3 \times 233 \). Total elements \( |A| = 233 \).
Elements in A are \( \{3, 6, 9, \dots, 699\} \).
Odd elements: \( \{3, 9, 15, \dots, 699\} \). This is an AP with \( a=3, d=6 \).
\( 699 = 3 + (n-1)6 \implies 696 = (n-1)6 \implies 116 = n-1 \implies n = 117 \).
Even elements: \( 233 - 117 = 116 \).
Set B: Multiples of 7 between 1 and 300.
Max element: \( 294 = 7 \times 42 \). Total elements \( |B| = 42 \).
Elements in B are \( \{7, 14, 21, \dots, 294\} \).
Odd elements: \( \{7, 21, 35, \dots, 287\} \). This is an AP with \( a=7, d=14 \).
\( 287 = 7 + (n-1)14 \implies 280 = (n-1)14 \implies 20 = n-1 \implies n = 21 \).
Even elements: \( 42 - 21 = 21 \).
Counting pairs \( a+b = \text{even} \):
Case 1: Both \( a, b \) are odd.
Ways \( = 117 \times 21 = 2457 \).
Case 2: Both \( a, b \) are even.
Ways \( = 116 \times 21 = 2436 \).
Total pairs without \( a \neq b \) condition: \( 2457 + 2436 = 4893 \).
Now handle \( a = b \):
This occurs when \( x \in A \cap B \), i.e., \( x \) is a multiple of \( \text{lcm}(3, 7) = 21 \).
Multiples of 21 in the overlapping range (up to 300):
\( 21, 42, 63, \dots, 294 \).
Number of such elements \( = 294 / 21 = 14 \).
For each such element, we have a pair \( (a, b) \) where \( a=b \). Since 14 is the total count, we subtract 14 from the previous total.
Order of C \( = 4893 - 14 = 4879 \).
Step 4: Final Answer:
The total number of valid pairs is 4879, which is option (A).