I gave the answer in a previous question without calculation. Here is the justiciation

• The probability a day has no people is $\left(1-\frac1n\right)^m$ so the expected number of days with no people is $n\left(1-\frac1n\right)^m$
• The probability a day has one person is ${m \choose 1}\frac1n\left(1-\frac1n\right)^{m-1}$ so the expected number of days with one person is $m\left(1-\frac1n\right)^{m-1}$
• So the expected number of days with two or more people is $$n - n \left(1-\frac1n\right)^m - m \left(1-\frac1n\right)^{m-1}$$

I gave the answer in a previous question without calculation. Here is the justification

• The probability a day has no people is $\left(1-\frac1n\right)^m$ so the expected number of days with no people is $n\left(1-\frac1n\right)^m$
• The probability a day has one person is ${m \choose 1}\frac1n\left(1-\frac1n\right)^{m-1}$ so the expected number of days with one person is $m\left(1-\frac1n\right)^{m-1}$
• So the expected number of days with two or more people is $$n - n \left(1-\frac1n\right)^m - m \left(1-\frac1n\right)^{m-1}$$