Question

how many arrangements of the digits $1,2,3,\ldots,9$ have the property that every digit (except the first) is no more than $3$ greater than the previous digit? (for example, the arrangement $214369578$ has this property. however, $312548697$ does not have the property, since $8$ occurs immediately after $4$, and $8>4 3$.)

Answer 1

The number of arrangements satisfying the given condition is 9 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 = 7,864. To find the number of arrangements of the digits 1,2,3,...,9 that have a specific property, use a systematic approach by considering each digit and its corresponding options. Multiply the number of options for each digit together to find the total number of arrangements.

Step-by-step explanation:

To find the number of arrangements of the digits 1,2,3,...,9 that have the property described, we can use a systematic approach:

1. Start with the first digit, which can be any of the digits 1 through 9. This gives us 9 options.
2. For the second digit, it can be no more than 3 greater than the first digit. So if the first digit is 1, the second digit can be 1, 2, 3, or 4. If the first digit is 2, the second digit can be 2, 3, 4, or 5. Continuing this pattern, there are 4 options for the second digit.
3. For the third digit, it can be no more than 3 greater than the second digit. So if the second digit is 1, the third digit can be 1, 2, 3, or 4. If the second digit is 2, the third digit can be 2, 3, 4, or 5. Continuing this pattern, there are again 4 options for the third digit.
4. And so on for each subsequent digit.

Since each digit has a fixed number of options, we can multiply the number of options for each digit to find the total number of arrangements.

Therefore, the number of arrangements satisfying the given condition is 9 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 = 7,864.