Key Concepts and Formulas

• Odd Number: A number is odd if its last digit is odd.
• Permutations with Repetitions: The number of permutations of $n$ objects, where $n_1$ are alike of one kind, $n_2$ are alike of another kind, ..., $n_k$ are alike of the $k^{th}$ kind (such that $n_1 + n_2 + \dots + n_k = n$), is given by: $\frac{n!}{n_1! n_2! \dots n_k!}$
• Sum Rule: If there are $m$ ways to do one thing and $n$ ways to do another, and the two things cannot be done at the same time, then there are $m + n$ ways to do one or the other.

Step-by-Step Solution

Step 1: Analyze the Digits and Identify Odd Digits

We are given the digits $1, 2, 2, 2, 3, 3, 5$. We need to form seven-digit odd numbers. For a number to be odd, its last digit must be odd. The odd digits in our set are $1, 3,$ and $5$. We will consider each of these as a separate case.

Step 2: Case 1: The Last Digit is 1

• Step 2.1: Fix the Last Digit We fix the last digit as 1: $\_ \_ \_ \_ \_ \_ 1$ Why? Fixing the last digit as 1 ensures that the number is odd.

• Step 2.2: Identify Remaining Digits The remaining digits are $2, 2, 2, 3, 3, 5$. Why? We started with $\{1, 2, 2, 2, 3, 3, 5\}$. Removing the 1 that we placed in the last digit leaves us with $\{2, 2, 2, 3, 3, 5\}$.

• Step 2.3: Calculate Permutations with Repetitions We have 6 digits to arrange, with the digit 2 repeated 3 times and the digit 3 repeated 2 times. The number of arrangements is: $\frac{6!}{3!2!}$ Why? We are arranging 6 objects, but we must account for the repetitions of the digits 2 and 3. The formula for permutations with repetitions handles this.

• Step 2.4: Compute the Result $\frac{6!}{3!2!} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(2 \times 1)} = \frac{720}{12} = 60$ There are 60 such numbers.

Step 3: Case 2: The Last Digit is 3

• Step 3.1: Fix the Last Digit We fix the last digit as 3: $\_ \_ \_ \_ \_ \_ 3$ Why? Fixing the last digit as 3 ensures that the number is odd and distinct from the previous case.

• Step 3.2: Identify Remaining Digits The remaining digits are $1, 2, 2, 2, 3, 5$. Why? We started with $\{1, 2, 2, 2, 3, 3, 5\}$. Removing the 3 that we placed in the last digit leaves us with $\{1, 2, 2, 2, 3, 5\}$.

• Step 3.3: Calculate Permutations with Repetitions We have 6 digits to arrange, with the digit 2 repeated 3 times. The number of arrangements is: $\frac{6!}{3!}$ Why? We are arranging 6 objects, but we must account for the repetition of the digit 2.

• Step 3.4: Compute the Result $\frac{6!}{3!} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1} = \frac{720}{6} = 120$ There are 120 such numbers.

Step 4: Case 3: The Last Digit is 5

• Step 4.1: Fix the Last Digit We fix the last digit as 5: $\_ \_ \_ \_ \_ \_ 5$ Why? Fixing the last digit as 5 ensures that the number is odd and distinct from the previous cases.

• Step 4.2: Identify Remaining Digits The remaining digits are $1, 2, 2, 2, 3, 3$. Why? We started with $\{1, 2, 2, 2, 3, 3, 5\}$. Removing the 5 that we placed in the last digit leaves us with $\{1, 2, 2, 2, 3, 3\}$.

• Step 4.3: Calculate Permutations with Repetitions We have 6 digits to arrange, with the digit 2 repeated 3 times and the digit 3 repeated 2 times. The number of arrangements is: $\frac{6!}{3!2!}$ Why? We are arranging 6 objects, but we must account for the repetitions of the digits 2 and 3.

• Step 4.4: Compute the Result $\frac{6!}{3!2!} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(2 \times 1)} = \frac{720}{12} = 60$ There are 60 such numbers.

Step 5: Calculate the Total Number of Odd Numbers

The three cases are mutually exclusive. Therefore, we add the results from each case to find the total number of seven-digit odd numbers: $60 + 120 + 60 = 240$

Common Mistakes & Tips

• Forgetting Repetitions: Always remember to divide by the factorials of the counts of repeated digits.
• Careless Digit Counting: Double-check which digits remain to be arranged in each case.
• Non-Mutually Exclusive Cases: Ensure your cases don't overlap. Here, a number cannot end in two different odd digits.

Summary

We found the number of seven-digit odd numbers that can be formed using the digits $1, 2, 2, 2, 3, 3, 5$ by considering three cases, one for each possible odd digit in the unit's place. We calculated the number of permutations with repetitions for each case and then summed the results to get the total.

The final answer is $\boxed{240}$.