Understanding the Problem: Counting Four-Digit Decimal Numbers

The question asks for the number of four-digit decimal numbers that are less than 1 and have no repeated digits. A decimal number less than 1 with four digits after the decimal point can be represented in the form $0.abcd$, where $a, b, c, d$ are digits.

The available digits we can use are $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$. The condition is that the four digits $a, b, c, d$ must all be distinct (no digit can be repeated). The total number of possible digits is 10.

Interpreting "Four-Digit Decimal Number"

The phrase "four digit decimal number" when referring to numbers less than 1 often implies that the first digit after the decimal point is non-zero. For example, $0.1234$ is clearly a four-digit decimal number. However, $0.0123$, while having four digits listed after the decimal, might sometimes be considered a three-digit decimal number ($0.0123 = 0.012300...$) in contexts where leading zeros after the decimal are ignored for determining the 'number' of digits. Given the options and the likely intended meaning in a permutation/combination context, we will proceed with the interpretation that the first digit after the decimal ($a$) cannot be zero.

Calculating the Number of Possibilities

We need to select four distinct digits for the positions $a, b, c,$ and $d$ such that $a \neq 0$. We will fill the positions one by one.

• For the first digit ($a$): This digit cannot be 0. So, we can choose from $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$. There are 9 possibilities for $a$.
• For the second digit ($b$): This digit must be different from $a$. We have 10 total digits $\{0, 1, ..., 9\}$. One digit has been used for position $a$. Since $a \neq 0$, the digit 0 is still available. So, we have $10 - 1 = 9$ remaining digits to choose from for $b$. There are 9 possibilities for $b$.
• For the third digit ($c$): This digit must be different from $a$ and $b$. Two distinct digits have been used for positions $a$ and $b$. From the 10 total digits, there are $10 - 2 = 8$ remaining digits to choose from for $c$. There are 8 possibilities for $c$.
• For the fourth digit ($d$): This digit must be different from $a, b,$ and $c$. Three distinct digits have been used for positions $a, b,$ and $c$. From the 10 total digits, there are $10 - 3 = 7$ remaining digits to choose from for $d$. There are 7 possibilities for $d$.

To find the total number of such four-digit decimal numbers with no repeated digits (under this interpretation), we multiply the number of possibilities for each position:

Total numbers = (Choices for $a$) $\times$ (Choices for $b$) $\times$ (Choices for $c$) $\times$ (Choices for $d$)

Total numbers = $9 \times 9 \times 8 \times 7$

Let's perform the multiplication:

$9 \times 9 = 81$

$81 \times 8 = 648$

$648 \times 7 = 4536$

So, there are 4536 such four-digit decimal numbers less than 1 with no repeated digits, under the assumption that the first digit after the decimal is non-zero.

Result

The total number of four-digit decimal numbers (<1) in which no digit is repeated, assuming the first digit after the decimal is not zero, is 4536.

Revision Table: Four-Digit Decimal Numbers

Additional Information: Permutations and Combinations

This problem involves the concept of permutations because the order of the digits matters (0.1234 is different from 0.4321). The number of permutations of $n$ distinct objects taken $r$ at a time is given by the formula:

$$P(n, r) = \frac{n!}{(n-r)!}$$

In our initial consideration, if we allowed the first digit after the decimal to be 0, we would be selecting and arranging 4 distinct digits from 10. This would be $P(10, 4) = \frac{10!}{6!} = 10 \times 9 \times 8 \times 7 = 5040$. However, the problem's phrasing and options suggest the constraint that the first digit after the decimal is non-zero. This extra constraint requires a step-by-step calculation as shown above, rather than a direct application of the standard permutation formula on all 10 digits for all four positions.