1. Digits given

We have the multiset of digits: $1,2,2,2,3,3,5$

We must form 7-digit odd numbers using all these digits.

So, the total number of valid numbers depends on the last digit being odd.

The odd digits available are: $1,3,3,5$

2. Casework based on the last digit

Since the number must be odd, the unit digit must be one of the odd digits.

Case 1: Last digit is $1$

Remaining digits are: $2,2,2,3,3,5$

Number of distinct arrangements of these 6 digits is: $\frac{6!}{3!2!} = \frac{720}{6\cdot 2} = 60$

Case 2: Last digit is $5$

Remaining digits are: $2,2,2,3,3,1$

Again, number of distinct arrangements is: $\frac{6!}{3!2!} = 60$

Case 3: Last digit is $3$

There are two 3's, but choosing either one gives the same type of remaining multiset: $1,2,2,2,3,5$

Now the repeated digit is only $2$ repeated 3 times. So number of distinct arrangements is: $\frac{6!}{3!} = \frac{720}{6} = 120$

3. Total number of odd numbers

Adding all cases: $60 + 60 + 120 = 240$

4. Final answer

The number of seven-digit odd numbers is: $\boxed{240}$

5. Comparison with stored answer

Stored correct answer = $240$

Our derived answer is also $240$, so it agrees.