The number of 4-digit integers in the closed interval [2022, 4482] formed by using the digits 0, 2,

Q: Why can't we just count all numbers starting with $2, 3, 4$ and subtract the ones that are too small or too big?

That is a valid strategy, but it is often more prone to error. In combinatorics, especially with restricted sets of digits, 'Constructive Counting' (building the numbers case-by-case) is safer. It forces you to respect the boundaries at every step, preventing the 'double-counting' or 'missing-counting' traps that subtraction methods often fall into.

Q: Wait, why did we count $4477$ as valid when the upper limit is $4482$?

This is a brilliant observation! The upper limit is $4482$, but our available digits are $\{0, 2, 3, 4, 6, 7\}$. We do not have the digits $8$ or $9$. Therefore, the largest number we can possibly form starting with $44$ is $4477$. Since $4477 < 4482$, every single combination starting with $44$ is automatically valid. You do not need to subtract anything!

Q: How do I know if repetition is allowed?

In JEE problems, if the question does not explicitly state 'distinct digits' or 'without repetition', you must assume repetition is allowed. Always look for that keyword. If it is absent, the 'slots' are independent, which actually makes the calculation easier!

Question

The number of 4-digit integers in the closed interval $[2022, 4482]$ formed by using the digits 0, 2, 3, 4, 6, 7 is _____________.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To solve for the number of four-digit integers in the range $[2022, 4482]$ using the set of digits $S = \{0, 2, 3, 4, 6, 7\}$, we must partition the problem based on the thousands digit.

The thousands digit must be $2, 3,$ or $4$ to remain within the interval. Choosing $6$ or $7$ would immediately exceed the upper bound of $4482$.

### Case One: Numbers Starting with $2$

Because the lower bound is $2022$, we must be precise to avoid values smaller than the threshold. We partition this into three sub-cases:

1. **Numbers of the form $202x$:** The first three digits are fixed as $2, 0, 2$. To satisfy the condition $\ge 2022$, the units digit $x$ must be chosen from $\{2, 3, 4, 6, 7\}$. This yields **$5$** valid numbers.

2. **Numbers of the form $20yx$ where $y > 2$:** Here, $y \in \{3, 4, 6, 7\}$ ($4$ choices) and $x \in \{0, 2, 3, 4, 6, 7\}$ ($6$ choices). This yields:
$$1 	imes 1 	imes 4 	imes 6 = 24 	ext{ numbers.}$$

3. **Numbers of the form $2zxx$ where $z > 0$:** Here, $z \in \{2, 3, 4, 6, 7\}$ ($5$ choices), while the tens and units digits can be any of the $6$ available digits. This yields:
$$1 	imes 5 	imes 6 	imes 6 = 180 	ext{ numbers.}$$

Summing these, the total for Case One is $5 + 24 + 180 = \mathbf{209}$.

### Case Two: Numbers Starting with $3$

Any number starting with $3$ is strictly between $2022$ and $4482$. Therefore, every combination is valid.

With $1$ choice for the thousands place and $6$ choices for each of the remaining three positions, we calculate:
$$1 	imes 6 	imes 6 	imes 6 = \mathbf{216} 	ext{ numbers.}$$

### Case Three: Numbers Starting with $4$

We must respect the upper bound of $4482$. We partition this into two logical segments:

1. **Numbers starting with $40xx, 42xx,$ or $43xx$:** The hundreds digit has $3$ choices $\{0, 2, 3\}$. The tens and units digits can be any of the $6$ digits. This yields:
$$1 	imes 3 	imes 6 	imes 6 = 108 	ext{ numbers.}$$

2. **Numbers starting with $44xx$:** The largest possible number we can form with our set is $4477$. Since $4477 < 4482$, all combinations starting with $44$ are valid. This yields:
$$1 	imes 1 	imes 6 	imes 6 = 36 	ext{ numbers.}$$

Summing these, the total for Case Three is $108 + 36 = \mathbf{144}$.

### Final Calculation

By summing the results of our three logical partitions, we arrive at the final count:

$$	ext{Total} = 209 + 216 + 144 = \mathbf{569}$$

The total number of four-digit integers that satisfy the given constraints is **$569$**.

Animated Solution for Mathematics - Permutations and Combinations: The number of 4-digit integers in the closed interval $[2022, 4482]$ formed by using the digits 0, 2, 3, 4, 6, 7 is _____________.

Enter Numerical Value:

Visualized Solution

Visual Anchor & Problem Setup

Strategy - The $4$ Slots

Case 1a: Numbers starting with $202$

Case 1b: Numbers starting with $203, 204, 206, 207$

Case 1c: Numbers starting with $22, 23, 24, 26, 27$

Total for Case 1

Case 2: Numbers starting with $3$

Case 3a: Numbers starting with $40, 42, 43$

Case 3b: The Upper Limit Trap

Final Computation

The Sigma Insight: Fundamental Principle of Counting

Analyzing the Setup

Case One

Case Two

Case Three

Final Calculation

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