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.