Natural numbers

Think carefully about how to count all natural numbers greater than 200 that are made up only of the digits 1, 2, 4, 6, and 8, with each of these digits appearing at most once in any given number.

Understanding the Problem

We need to find how many natural numbers greater than 200 can be formed using only the digits from the set {1, 2, 4, 6, 8}, where each digit in this set appears at most once in any number.

Important Notes:
1. Digits Used: Only 1, 2, 4, 6, and 8 can be used. No other digits (like 0, 3, 5, 7, 9) are allowed.
2. No Repeats: Each of these digits can appear at most once in a number. So, numbers like 224 or 121 are invalid because digits 2 and 1 repeat, respectively.
3. Greater than 200: The numbers must be greater than 200. This affects the possible numbers based on their digit length.

Given that we can only use the digits {1, 2, 4, 6, 8}, and each digit can be used at most once, the numbers we can form will have lengths from 1 digit up to 5 digits (since there are only 5 unique digits available).

However, since the numbers must be greater than 200, let's consider the possible lengths:

- 3-digit numbers: These range from 100 to 999 in general, but with our digits, the smallest is 124 and the largest is 864 (or similar permutations). Many of these will be greater than 200.
- 4-digit numbers: All 4-digit numbers formed with our digits will be ≥ 1246, which is certainly > 200.
- 5-digit numbers: Similarly, all 5-digit numbers formed will be > 200.

Given that, we need to count all 3-digit, 4-digit, and 5-digit numbers formed under the given constraints.

Counting 3-digit Numbers > 200

For a 3-digit number to be greater than 200, its hundreds digit must be 2, 4, 6, or 8 (since 1 would make it 1xx, which is ≤ 199 < 200).

Digits Available: {1, 2, 4, 6, 8}

Hundreds digit (first digit): Must be one of {2, 4, 6, 8} → 4 choices.

Tens digit: Can be any of the remaining 4 digits (since one is already used as the hundreds digit).

Units digit: Can be any of the remaining 3 digits.

So, total 3-digit numbers = 4 (hundreds) × 4 (tens) × 3 (units) = 48.

For 4-digit numbers, all are > 200 (since the smallest is 1246 > 200).

We have 5 digits total, using 4 at a time with no repeats.

Number of 4-digit numbers = number of permutations of 5 digits taken 4 at a time = 5P4 = 5! / (5-4)! = 5 × 4 × 3 × 2 = 120.

Counting 5-digit Numbers

For 5-digit numbers, we use all 5 digits exactly once (since no repeats and we have exactly 5 digits).

Number of 5-digit numbers = number of permutations of 5 digits = 5! = 120.

Can we have 6-digit numbers? No, because we only have 5 unique digits, and we can't repeat any to form a 6-digit number under the given constraints.

Total Count

Now, sum up all valid numbers:

- 3-digit: 48
- 4-digit: 120
- 5-digit: 120

Total = 48 + 120 + 120 = 288.