Key Concepts and Formulas

• Permutations with Repetitions: The number of distinct permutations of $n$ objects, where $n_1$ are of one type, $n_2$ are of a second type, ..., $n_k$ are of a $k$-th type (and $n_1 + n_2 + ... + n_k = n$), is given by $\frac{n!}{n_1! n_2! ... n_k!}$.
• Combinations: The number of ways to choose $r$ objects from a set of $n$ distinct objects is given by $\binom{n}{r} = \frac{n!}{r!(n-r)!}$.
• The problem requires understanding that since we have 6 digits to fill and only 5 distinct digits to use, one of the digits must be repeated.

Step-by-Step Solution

Step 1: Determine the Repeated Digit

We need to form a 6-digit number using the digits {1, 3, 5, 7, 9}. Since we have only 5 distinct digits and need to form a 6-digit number, one of the digits must be repeated. Our task is to determine which of the 5 digits will be repeated.

Step 2: Choose the Digit to be Repeated

Out of the five digits (1, 3, 5, 7, 9), we need to choose one digit to be repeated. The number of ways to choose one digit out of five is given by the combination formula: $\binom{5}{1} = \frac{5!}{1!(5-1)!} = \frac{5!}{1!4!} = 5$ This means we have 5 choices for the digit that will be repeated.

Step 3: Form the Set of 6 Digits

Once we've chosen the repeating digit, we have a set of 6 digits where one digit appears twice, and the other four digits appear once. For example, if we choose '1' as the repeating digit, our set is {1, 1, 3, 5, 7, 9}.

Step 4: Arrange the Six Digits

Now, we need to arrange these 6 digits to form a 6-digit number. We have a total of 6 digits, with one digit repeating twice. Using the formula for permutations with repetitions, we have: $\frac{6!}{2! 1! 1! 1! 1!} = \frac{6!}{2!} = \frac{720}{2} = 360$ The $6!$ accounts for all possible arrangements if all digits were distinct, and the $2!$ corrects for the overcounting due to the two identical digits.

Step 5: Calculate the Total Number of 6-Digit Numbers

To find the total number of such 6-digit numbers, we multiply the number of ways to choose the repeating digit (from Step 2) by the number of ways to arrange the resulting set of six digits (from Step 4): $\text{Total numbers} = (\text{Ways to choose repeating digit}) \times (\text{Ways to arrange the 6 digits})$ $\text{Total numbers} = 5 \times \frac{6!}{2!}$ $\text{Total numbers} = 5 \times \frac{720}{2} = 5 \times 360 = 1800$ We can rewrite this as: $\text{Total numbers} = \frac{5}{2} (6!) = \frac{5}{2} (720) = 1800$

Common Mistakes & Tips:

• Forgetting to Choose the Repeating Digit: A common mistake is to directly arrange the 6 digits with one repetition without considering which of the 5 digits is repeated.
• Incorrect Formula Application: Ensure proper use of combinations for selection and permutations with repetitions for arrangement.
• Misinterpreting "Only and All": The phrase "only and all the five digits" emphasizes that no other digits can be used, and all of 1, 3, 5, 7, and 9 must appear at least once, implying one digit is repeated.

Summary

The problem involves forming a 6-digit number using only the digits 1, 3, 5, 7, and 9, with each digit appearing at least once. This means one digit must be repeated. We first choose which digit is repeated using combinations, and then arrange the resulting 6 digits using permutations with repetitions. The total number of such 6-digit numbers is calculated by multiplying the number of ways to choose the repeated digit by the number of ways to arrange the digits, which results in $\frac{5}{2}(6!)$.

The final answer is $\boxed{\frac{5}{2}(6!)}$, which corresponds to option (A).