The number of seven digits odd numbers, that can be formed using all the seven digits 1, 2, 2, 2, 3, 3, 5 is ______.

We need to find the number of seven-digit odd numbers that can be formed using all the digits 1, 2, 2, 2, 3, 3, 5.

A number is odd if its last digit is odd. The odd digits available here are 1, 3, and 5.

If the last digit is 1, the remaining digits are 2, 2, 2, 3, 3, 5. The number of arrangements of these six digits is $$\dfrac{6!}{3! \cdot 2!} = \dfrac{720}{12} = 60$$.

If the last digit is 3, the remaining digits are 1, 2, 2, 2, 3, 5. In this case there are three 2’s and one each of 1, 3, and 5, so the number of ways to arrange them is $$\dfrac{6!}{3!} = \dfrac{720}{6} = 120$$.

If the last digit is 5, the remaining digits are 1, 2, 2, 2, 3, 3. With three 2’s, two 3’s, and one 1, the count is $$\dfrac{6!}{3! \cdot 2!} = \dfrac{720}{12} = 60$$.

Adding these cases gives $$60 + 120 + 60 = 240$$. Therefore, the total number of such seven-digit odd numbers is $$\boxed{240}$$.