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A set of $n$ ($n \geq 7$) people is given. What is the probability that their birthdays cover all days of the week?

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    If we assume equal probabilities and independence, this problem is closely related to the well-known coupon-collector-problem. – 2017-02-08

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Use the principle of inclusion and exclusion $$7^n-\binom{7}{1}6^n+\binom{7}{2}5^n+...=\sum_{i=0}^{7}(-1)^i\binom{7}{i}(7-i)^n$$ This is the total number of possibilities, so for probability you must divide it be $7^n$

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    can u please explain the reasoning behind this answer? – 2017-02-08
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    @Kiran Assume there is not any restriction, and each person can select any day, so totally we have $7^n$ different possibility, now there is the possibility that one day is not chosen at all, then there are 6 possibility for each person. We have extracted the case that there are more than 1 day which are missed extra, so we must add it, ... – 2017-02-08
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    This is the number of possibilities, for probability, you must divide it by the total number of situations which is $7^n$ – 2017-02-08
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There are $7!$ orders of the days of the week and $7^7$ ways to pick the birthdays, so $\frac {7!}{7^7}=\frac {720}{117649}\approx 0.006$ Another way to see it is to note that we must avoid any duplicate days. The first person cannot duplicate an existing day. The second will not duplicate the first with probability $\frac 67$. Assuming the second does not match the first, the third will not match either with probability $\frac 57$ and so on. This also gives $\frac {7!}{7^7}$

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    it is valid for $n<7$ also right? – 2017-02-08
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    @Kiran: If you mean the pattern would be the same if you had five characteristics and five people, yes, it would be $\frac {5!}{5^5}$ – 2017-02-08
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    what if $5$ people and $7$ days? still, $7!$ order for days and $7^5$ ways to pick the birthdays and therefore $5!/7^5$, am i correct? – 2017-02-08
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    @Kiran: With five people and seven days you cannot account for them all, so the chance is zero. – 2017-02-08
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    It is wrong, as when $n$ goes to infinity, the probability must go toward $1$ – 2017-02-08
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    @CLAUDE: My solution relies on the fact that the number of people is the same as the number of possible birthdays. As such as the number goes to infinity the probability must go to zero, which it does. If there are a fixed number of birthdays and the number of people goes to infinity the probability must go to 1, but I do not address that. – 2017-02-08