Count of Four Digit Integers in Given Closed Interval

To find the number of 4-digit integers in the closed interval $[2022, 4482]$ formed using the set of digits $S = \{0, 2, 3, 4, 6, 7\}$, we consider a general four-digit integer represented as $N = d_1 d_2 d_3 d_4$, where $d_1 \in \{2, 3, 4\}$ and $d_2, d_3, d_4 \in S$.

Since there is no restriction stating that digits cannot be repeated, repetition of digits is allowed. We break the problem down into cases based on the thousands digit $d_1$.

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Case 1: $d_1 = 2$

We require $N = 2 d_2 d_3 d_4 \ge 2022$.

1. Subcase 1.1: $d_2 \in \{2, 3, 4, 6, 7\}$ (5 choices)

   • Since $d_2 \ge 2$, $N \ge 2200 > 2022$.
   • $d_3$ and $d_4$ can each be any of the $6$ digits in $S$.
   • Number of integers = $5 \times 6 \times 6 = 180$.

2. Subcase 1.2: $d_2 = 0$

   • We need $20d_3d_4 \ge 2022 \implies d_3d_4 \ge 22$.
   • If $d_3 \in \{3, 4, 6, 7\}$ (4 choices): $d_4$ can be any of the $6$ digits in $S$. Number of integers = $4 \times 6 = 24$.
   • If $d_3 = 2$: We need $202d_4 \ge 2022 \implies d_4 \ge 2$. $d_4 \in \{2, 3, 4, 6, 7\}$ (5 choices). Number of integers = $5$.
   • If $d_3 = 0$: The maximum value is $2007 < 2022$, so there are $0$ valid integers.

Total for Case 1 = $180 + 24 + 5 = 209$.

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Case 2: $d_1 = 3$

• Any four-digit number starting with $3$ formed using elements of $S$ satisfies $3000 \le N \le 3777$, which lies completely within $[2022, 4482]$.
• $d_2, d_3,$ and $d_4$ can each be chosen from $S$ in $6$ ways.

Total for Case 2 = $1 \times 6 \times 6 \times 6 = 216$.

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Case 3: $d_1 = 4$

We require $N = 4 d_2 d_3 d_4 \le 4482$.

1. Subcase 3.1: $d_2 \in \{0, 2, 3\}$ (3 choices)

   • Since $d_2 \le 3$, $N \le 4377 < 4482$.
   • $d_3$ and $d_4$ can each be chosen from $S$ in $6$ ways.
   • Number of integers = $3 \times 6 \times 6 = 108$.

2. Subcase 3.2: $d_2 = 4$

   • The largest number starting with $44$ using digits from $S$ is $4477$.
   • Since $4477 \le 4482$, all choices for $d_3$ and $d_4$ are valid.
   • Number of integers = $1 \times 6 \times 6 = 36$.

3. Subcase 3.3: $d_2 \in \{6, 7\}$

   • The smallest number starting with $46$ is $4600 > 4482$, so there are $0$ valid integers.

Total for Case 3 = $108 + 36 = 144$.

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Total Count

Adding the results from all cases: $\text{Total number of integers} = 209 + 216 + 144 = 569$