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_________.

The correct answer is 762.

Let us solve the problem step-by-step using the Principle of Inclusion-Exclusion.

A seven-digit number is formed using the digits 0, 1, and 2. Since it is a seven-digit number, the first digit (the most significant digit) cannot be 0. Thus, the first digit must be either 1 or 2.

Let $A$ be the set of seven-digit numbers in $S$ where the digit 0 appears exactly twice.
Let $B$ be the set of seven-digit numbers in $S$ where the digit 1 appears exactly twice.

We need to find the number of elements in $S$ such that at least one of the digits 0 and 1 appears exactly twice. This is represented by the cardinality of the union of $A$ and $B$:
$n(A\cup B)=n\left(A\right)+n\left(B\right)-n(A\cap B)$

Step 1: Finding $n\left(A\right)$ (0 appears exactly twice)
Since 0 cannot be the first digit, the first digit must be 1 or 2. The remaining 6 positions must contain exactly two 0s. The other 4 positions can be filled with either 1 or 2.

We consider two cases based on the first digit:
Case 1: The first digit is 1.
We choose 2 positions for 0 from the remaining 6 positions in $C^6$₂=15 ways.
The remaining 4 positions can be filled with 1 or 2 in $2^4=16$ ways.
Number of ways = $15\times 16=240$.

Case 2: The first digit is 2.
Similarly, we choose 2 positions for 0 from the remaining 6 positions in $C^6$₂=15 ways.
The remaining 4 positions can be filled with 1 or 2 in $2^4=16$ ways.
Number of ways = $15\times 16=240$.

Therefore, the total number of ways is:
$n\left(A\right)=240+240=480$

Step 2: Finding $n\left(B\right)$ (1 appears exactly twice)
The first digit can be 1 or 2. We consider two cases based on the first digit:

Case 1: The first digit is 1.
Since the first digit is 1, and 1 must appear exactly twice overall, we need to place exactly one more 1 in the remaining 6 positions.
We choose 1 position for the second 1 in $C^6$₁=6 ways.
The remaining 5 positions can be filled with 0 or 2 in $2^5=32$ ways.
Number of ways = $6\times 32=192$.

Case 2: The first digit is 2.
Since the first digit is 2, the remaining 6 positions must contain exactly two 1s.
We choose 2 positions for the 1s in $C^6$₂=15 ways.
The remaining 4 positions can be filled with 0 or 2 in $2^4=16$ ways.
Number of ways = $15\times 16=240$.

Therefore, the total number of ways is:
$n\left(B\right)=192+240=432$

Step 3: Finding $n(A\cap B)$ (both 0 and 1 appear exactly twice)
In this case, the number must contain exactly two 0s and exactly two 1s. The remaining three digits must be 2. The first digit can be 1 or 2.

Case 1: The first digit is 1.
We need one more 1 and two 0s in the remaining 6 positions.
First, choose 1 position out of 6 for the 1: $C^6$₁=6 ways.
Then, choose 2 positions out of the remaining 5 for the 0s: $C^5$₂=10 ways.
The remaining 3 positions are automatically filled with 2s.
Number of ways = $6\times 10=60$.

Case 2: The first digit is 2.
We need two 1s and two 0s in the remaining 6 positions.
First, choose 2 positions out of 6 for the 1s: $C^6$₂=15 ways.
Then, choose 2 positions out of the remaining 4 for the 0s: $C^4$₂=6 ways.
The remaining 2 positions are automatically filled with 2s.
Number of ways = $15\times 6=90$.

Therefore, the total number of ways is:
$n(A\cap B)=60+90=150$

Step 4: Applying the Principle of Inclusion-Exclusion
We now compute the final count:
$n(A\cup B)=480+432-150$
$n(A\cup B)=912-150=762$

Thus, there are exactly 762 such seven-digit numbers.