This question assumes that a year has 365 days, and each day of the year is equally likely to be a birthday for someone.
At first I thought I simply had to select a day and then 2 individuals ($\frac{\binom{10}{2}}{365}$, but then this can't make sense because I am seeking how many days, not people.
Best Answer
The probability a given day has no birthdays is $\left(1-\frac{1}{365}\right)^{10}$ so the expected number of days with no birthday is $365$ times this, about $355.1224$.
The probability a given day has exactly one birthday is $10 \times \frac{1}{365} \times \left(1-\frac{1}{365}\right)^9$ so the expected number of days with exactly one birthday is $365$ times this, about $9.7561$.
Subtract these two figures from $365$ and you get a result of about $0.1215$ for the expected number of days with two or more birthdays.