We need to count seven-digit numbers formed from digits $0,1,2$ such that:

• the first digit is not $0$,
• and at least one of the digits $0$ and $1$ appears exactly twice.

Let:

• $A$ = set of valid seven-digit numbers where digit $0$ appears exactly twice,
• $B$ = set of valid seven-digit numbers where digit $1$ appears exactly twice.

We want: $|A \cup B| = |A| + |B| - |A \cap B|.$

1. Count $|A|$: digit $0$ appears exactly twice

A 7-digit number cannot start with $0$, so the two zeroes must be placed among the last 6 positions.

Number of ways to choose positions of the two zeroes: $\binom{6}{2} = 15.$

The remaining 5 positions can each be filled with either $1$ or $2$: $2^5 = 32.$

Hence, $|A| = \binom{6}{2} \cdot 2^5 = 15 \cdot 32 = 480.$

2. Count $|B|$: digit $1$ appears exactly twice

Now we count 7-digit numbers where exactly two digits are $1$.

Since $1$ can appear anywhere, first choose the 2 positions for the digit $1$: $\binom{7}{2} = 21.$

The remaining 5 positions are to be filled with $0$ or $2$, but the first digit of the whole number cannot be $0$.

So we split into cases.

Case 1: First position is one of the chosen positions for digit $1$

Then the first digit is automatically nonzero.

Choose the other position of $1$ from the remaining 6 positions: $\binom{6}{1} = 6.$

The remaining 5 positions can be filled with $0$ or $2$ freely: $2^5 = 32.$

Count: $6 \cdot 32 = 192.$

Case 2: First position is not a position of digit $1$

Then both $1$'s are among the last 6 positions: $\binom{6}{2} = 15.$

Now among the remaining 5 positions, the first digit must be $2$ (cannot be $0$), and the other 4 positions can be $0$ or $2$: $1 \cdot 2^4 = 16.$

Count: $15 \cdot 16 = 240.$

Thus, $|B| = 192 + 240 = 432.$

3. Count $|A \cap B|$: both $0$ and $1$ appear exactly twice

We need numbers with:

• exactly two $0$'s,
• exactly two $1$'s,
• therefore the remaining three digits are $2$'s.

Again, since the first digit cannot be $0$, count by cases.

Case 1: First digit is $1$

Then we need:

• one more $1$ among the remaining 6 positions,
• two $0$'s among the remaining 6 positions,
• the rest are $2$'s.

Choose position of the second $1$: $\binom{6}{1} = 6.$

Choose positions of the two zeroes from remaining 5 positions: $\binom{5}{2} = 10.$

Count: $6 \cdot 10 = 60.$

Case 2: First digit is $2$

Then among the remaining 6 positions, choose:

• two positions for $0$,
• two positions for $1$,
• remaining two are $2$.

Choose positions of two zeroes: $\binom{6}{2} = 15,$ then positions of two ones from remaining 4: $\binom{4}{2} = 6.$

Count: $15 \cdot 6 = 90.$

Therefore, $|A \cap B| = 60 + 90 = 150.$

4. Apply inclusion-exclusion

$|A \cup B| = |A| + |B| - |A \cap B|$ $= 480 + 432 - 150$ $= 762.$

Final Answer

The required number of elements is $\boxed{762}.$

This matches the stored correct answer.