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 the digits 0 and 1 appears exactly twice in \(x\), is equal to ______.

Let \(\mathrm{A} \rightarrow\) " 0 " appear exactly twice.
and \(\mathrm{B} \rightarrow " 1 "\) appear exactly twice.
\(\therefore \mathrm{A} \cap \mathrm{B} \rightarrow\) " 0 " and " 1 " both appears exactly twice.
\(\mathrm{n}(\mathrm{A})=-\underbrace{-------}\)
\(={ }_{\text {placing zero }}^6 \mathrm{C}_2(1)(2)^5=\frac{6 \times 5}{2} \times 2^5=480\) for \(\mathrm{n}(\mathrm{B})\)
C-I : \(1\) at first place
\(\underline{1}\underbrace{-------}\)
\(\text { Number of ways }=\underset{\text { placing } 1}{} \mathrm{}^6{C}_1(1)(2)^5=192\)
C-II : 2 at first place
\(\underline{2}\underbrace{-------}\)
Number of ways \(=\underset{\text { placing } 1}{{ }^6 \mathrm{C}_2(1)(2)^4}=\frac{6 \times 5}{2} \times 2^4=240\)
\(n(B)=240+192\)
for \(\mathrm{n}(\mathrm{A} \cap \mathrm{B})\)
\(\underline{}\underbrace{-------}\)
for \(\mathrm{n}(\mathrm{A} \cap \mathrm{B})=\underset{\text { placing zero }}{{ }^6 \mathrm{C}_2(1)} \times \underset{\text { placing } 1}{{ }^5 \mathrm{C}_2(1)} \times \underset{\text { 2at rest places } 1}{(1)}\) \(=\frac{6 \times 5}{2} \times \frac{5 \times 4}{2}=150\)
\(\begin{aligned} & \therefore \mathrm{n}(\mathrm{A} \cup \mathrm{B})=\mathrm{n}(\mathrm{A})+\mathrm{n}(\mathrm{B})-\mathrm{n}(\mathrm{A} \cap \mathrm{B}) \\ & =480+(192+240)-150 \\ & =762\end{aligned}\)