Let $S$ be the set of all seven-digit numbers that can be formed using the digits $0$, $1$ and $2$. For example, $2210222$ is in $S$, but $0210222$ is NOT in $S$. Then the number of elements $x$ in $S$ such that at least one of the digits $0$ and $1$ appears exactly twice in $x$, is equal to ______.

$\text{Given: seven-digit numbers using only the digits } 0, 1, 2$ $\text{leading digit } \neq 0$ $\text{To find: the number of } x \in S \text{ with at least one of } 0, 1 \text{ appearing exactly twice}$ $\text{i.e. count } x \text{ with } 0 \text{ twice, or } 1 \text{ twice, or both twice}$ $A = \{ x \in S : 0 \text{ appears exactly twice} \}, \quad B = \{ x \in S : 1 \text{ appears exactly twice} \}$ $|A \cup B| = |A| + |B| - |A \cap B|$ $\text{Total elements of } S = 2 \cdot 3^{6} = 1458$ $\text{0 twice: choose 2 of the 6 non-leading positions } \binom{6}{2}$ $\text{remaining 5 positions are } 1 \text{ or } 2: \ 2^{5}$ $|A| = \binom{6}{2} \cdot 2^{5} = 15 \cdot 32 = 480$ $\text{Leading } = 1: \ \binom{6}{1} \cdot 2^{5} = 6 \cdot 32 = 192$ $\text{Leading } = 2: \ \binom{6}{2} \cdot 2^{4} = 15 \cdot 16 = 240$ $|B| = 192 + 240 = 432$ $\text{Leading } = 1: \ \binom{6}{1} \cdot \binom{5}{2} = 6 \cdot 10 = 60$ $\text{Leading } = 2: \ \binom{6}{2} \cdot \binom{4}{2} = 15 \cdot 6 = 90$ $|A \cap B| = 60 + 90 = 150$ $|A \cup B| = 480 + 432 - 150 = 762$ $\boxed{762}$