<s>[INST] In a room of $n \geq 7$ people, what is the probability that, for each day of the week, there is at least one person whose birthday falls on that day? In a room of $n \geq 7$ people, what is the probability that, for each day of the week, there is at least one person whose birthday falls on that day? Assume each day of the week has equal probability.
I am thinking about allocating each day one person and then stars and bars the rest? But cant think of the sample space.
I also consider using complement:
$$1- \binom{7}{1}\left(\frac{6}{7}\right)^n - \binom{7}{2}\left(\frac{5}{7}\right)^n \ldots$$
but then I will double count. I tried inclusion-exclusion, but I can't figure it out.
 [/INST] Since there are seven possible days of the week on which each person could have been born, there are $7^n$ possible birthday assignments.
If at least one person is born on each day of the week, we must exclude those birthday assignments in which at least one day is missing.  There are $\binom{7}{k}$ ways to exclude $k$ days of the week and $(7 - k)^n$ ways to distribute the $n$ birthdays to the remaining $7 - k$ days of the week.  Hence, by the Inclusion-Exclusion Principle, the number of favorable cases is
$$\sum_{k = 0}^{7} (-1)^k\binom{7}{k}(7 - k)^n = 7^n - \binom{7}{1}6^n + \binom{7}{2}5^n - \binom{7}{3}4^n + \binom{7}{4}3^n - \binom{7}{5}2^n + \binom{7}{6}1^n - \binom{7}{7}0^n$$
Hence, the probability that at least one person is born on each day of the week is
$$\frac{7^n - \dbinom{7}{1}6^n + \dbinom{7}{2}5^n - \dbinom{7}{3}4^n + \dbinom{7}{4}3^n - \dbinom{7}{5}2^n + \dbinom{7}{6}1^n - \dbinom{7}{7}0^n}{7^n}$$
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