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 ________.

Let a seven-digit number $$x$$ contain $$a_0$$ zeros, $$a_1$$ ones and $$a_2$$ twos, so that

$$a_0 + a_1 + a_2 = 7$$ and $$a_0,\,a_1,\,a_2 \ge 0$$.

Because the first (left-most) digit cannot be $$0$$, we count the admissible arrangements for each triple $$(a_0,a_1,a_2)$$ as follows.

Total arrangements with these counts = $$\dfrac{7!}{a_0!\,a_1!\,a_2!}$$.
Arrangements whose first digit is $$0$$: fix a zero at the first place and permute the remaining $$6$$ digits, giving $$\dfrac{6!}{(a_0-1)!\,a_1!\,a_2!}$$ (this term appears only when $$a_0\ge 1$$).

Hence the required number for the triple $$(a_0,a_1,a_2)$$ is

$$N(a_0,a_1,a_2)= \begin{cases} \dfrac{7!}{a_0!\,a_1!\,a_2!}-\dfrac{6!}{(a_0-1)!\,a_1!\,a_2!}, & a_0\ge 1\\[6pt] \dfrac{7!}{a_0!\,a_1!\,a_2!}, & a_0=0 \end{cases}$$

For $$a_0\ge 1$$, simplifying gives

$$N(a_0,a_1,a_2)=\dfrac{6!}{a_0!\,a_1!\,a_2!}\,(7-a_0)\qquad -(1)$$

The problem asks for the count of all numbers in which at least one of the digits $$0$$ or $$1$$ appears exactly twice. We split this into three disjoint cases.

Let the possible values of $$a_1$$ be $$0,1,3,4,5$$ (they must stay within $$0\le a_1\le 5$$). For each value we compute $$a_2=7-2-a_1$$ and use $$(1)$$.

$$\begin{aligned} a_1=0,&\;a_2=5:&\;N=\dfrac{6!\,(7-2)}{2!\,0!\,5!}=15\\ a_1=1,&\;a_2=4:&\;N=\dfrac{6!\,(7-2)}{2!\,1!\,4!}=75\\ a_1=3,&\;a_2=2:&\;N=\dfrac{6!\,(7-2)}{2!\,3!\,2!}=150\\ a_1=4,&\;a_2=1:&\;N=\dfrac{6!\,(7-2)}{2!\,4!\,1!}=75\\ a_1=5,&\;a_2=0:&\;N=\dfrac{6!\,(7-2)}{2!\,5!\,0!}=15 \end{aligned}$$

Total for Case 1 = $$15+75+150+75+15 = 330$$.

Now $$a_0$$ can be $$0,1,3,4,5$$. Compute $$a_2=7-a_0-2$$ and apply the appropriate formula.

$$\begin{aligned} a_0=0,&\;a_2=5:&\;N=\dfrac{7!}{0!\,2!\,5!}=21\\ a_0=1,&\;a_2=4:&\;N=\dfrac{6!\,(7-1)}{1!\,2!\,4!}=90\\ a_0=3,&\;a_2=2:&\;N=\dfrac{6!\,(7-3)}{3!\,2!\,2!}=120\\ a_0=4,&\;a_2=1:&\;N=\dfrac{6!\,(7-4)}{4!\,2!\,1!}=45\\ a_0=5,&\;a_2=0:&\;N=\dfrac{6!\,(7-5)}{5!\,2!\,0!}=6 \end{aligned}$$

Total for Case 2 = $$21+90+120+45+6 = 282$$.

Both digits $$0$$ and $$1$$ appear exactly twice.

$$N=\dfrac{6!\,(7-2)}{2!\,2!\,3!}=150$$

Summing all three cases:

$$330 + 282 + 150 = 762$$

Therefore, the required number of seven-digit numbers is
762.