Seven Digit Numbers with Repeated Digits Zero and One

To find the number of elements in $S$ such that at least one of the digits $0$ and $1$ appears exactly twice, we use the Principle of Inclusion-Exclusion.

Let:

• $S$ be the set of all $7$-digit numbers formed using the digits $0, 1, 2$. The first digit $d_1 \in \{1, 2\}$, and the remaining six digits $d_2, d_3, \dots, d_7 \in \{0, 1, 2\}$.
• $A$ be the set of numbers in $S$ where the digit $0$ appears exactly twice.
• $B$ be the set of numbers in $S$ where the digit $1$ appears exactly twice.

We wish to find $|A \cup B| = |A| + |B| - |A \cap B|$.

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Step 1: Calculate $|A|$ (Digit $0$ appears exactly twice)

Since the first digit $d_1 \neq 0$, both $0$'s must appear in two of the remaining $6$ positions:

• Choose $2$ positions out of $6$ for the digit $0$: $\binom{6}{2} = 15$ ways.
• The first digit $d_1$ can be chosen from $\{1, 2\}$: $2$ choices.
• The remaining $4$ positions can each be chosen from $\{1, 2\}$: $2^4 = 16$ choices.

$|A| = \binom{6}{2} \times 2 \times 2^4 = 15 \times 2 \times 16 = 480$

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Step 2: Calculate $|B|$ (Digit $1$ appears exactly twice)

We consider two disjoint cases based on the first digit $d_1$:

1. Case 1: $d_1 = 1$

   • The remaining one digit $1$ must be placed in one of the other $6$ positions: $\binom{6}{1} = 6$ ways.
   • The remaining $5$ positions are filled using $\{0, 2\}$: $2^5 = 32$ ways. $\text{Ways} = 1 \times 6 \times 32 = 192$

2. Case 2: $d_1 = 2$

   • Both digits $1$ must be placed in two of the other $6$ positions: $\binom{6}{2} = 15$ ways.
   • The remaining $4$ positions are filled using $\{0, 2\}$: $2^4 = 16$ ways. $\text{Ways} = 1 \times 15 \times 16 = 240$

$|B| = 192 + 240 = 432$

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Step 3: Calculate $|A \cap B|$ (Both digits $0$ and $1$ appear exactly twice)

The digit $0$ must occupy $2$ of the last $6$ positions: $\binom{6}{2} = 15$ ways.

1. Case 1: $d_1 = 1$

   • Digit $1$ must occupy $1$ of the remaining $4$ positions: $\binom{4}{1} = 4$ ways.
   • The other $3$ positions are filled with digit $2$: $1^3 = 1$ way. $\text{Ways} = 15 \times 4 \times 1 = 60$

2. Case 2: $d_1 = 2$

   • Digit $1$ must occupy $2$ of the remaining $4$ positions: $\binom{4}{2} = 6$ ways.
   • The other $2$ positions are filled with digit $2$: $1^2 = 1$ way. $\text{Ways} = 15 \times 6 \times 1 = 90$

$|A \cap B| = 60 + 90 = 150$

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Step 4: Combine using the Principle of Inclusion-Exclusion

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

The number of elements in $S$ satisfying the given condition is $762$.