Suppose $A$ is denoted the set of all numbers between 1 and 700 which are divisible by 3 and let $B$ is denoted the set of all numbers between 1 and 300 which are divisible by 7. If $C = \{(a, b) | a \in A, b \in B, a \neq b \text{ and } a + b = \text{even number}\}$, then order of $C$ is

The set $A$ contains multiples of $3$ from $1$ to $700$.
Number of elements in $A = \lfloor \dfrac{700}{3} \rfloor = 233$.
Number of even elements in $A = \lfloor \dfrac{700}{6} \rfloor = 116$.
Number of odd elements in $A = 233 - 116 = 117$.

The set $B$ contains multiples of $7$ from $1$ to $300$.
Number of elements in $B = \lfloor \dfrac{300}{7} \rfloor = 42$.
Number of even elements in $B = \lfloor \dfrac{300}{14} \rfloor = 21$.
Number of odd elements in $B = 42 - 21 = 21$.

For the sum $a + b$ to be even, both $a$ and $b$ must be even, or both must be odd.
Total number of such pairs $(a, b)$ is $(116 \times 21) + (117 \times 21) = 2436 + 2457 = 4893$.

The condition $a \neq b$ requires subtracting pairs where $a = b$.
If $a = b$, then $a$ is a multiple of both $3$ and $7$, which means $a$ is a multiple of $21$.
Since $b \in B$, we have $b \le 300$, so $a = b \le 300$.
Number of such pairs is $\lfloor \dfrac{300}{21} \rfloor = 14$.
Since $a = b$ makes $a + b = 2a$ which is even, these $14$ pairs are included in the $4893$ pairs.

The order of $C$ is $4893 - 14 = 4879$.

Answer: 4879