# How many arrangements can a team of 8 students have such at all of them have their birthdays in 2 days (but not all on one day)

The full question is from:

Find the probability that 8 students in a team will all have their birthday on exactly two days of the week (but not all in one day)?

However, the answers didn't address the main problem, which is finding the total probability. The $\Omega$ is $7^8$ for obvious reasons. So, to find the probability that 8 students in a team have all their birthdays on 2 days (but not all on one day) I thought of this:

Consider: Number of possibilities for 8 students having their birthday all on Sat. or Sun.

$\Omega($8 students have all their birthdays on say Saturday and Sunday (but not all on one day))

Then first of all we have these possibly combinations of Saturday an Sunday:

• 7 on Saturday, 1 on Sunday
• 6 on Saturday, 2 on Sunday
• ...... 1 on Saturday, 7 on Sunday

There are 7 possible "combinations" of how to arrange the players to satisfy the condition that all have their birthdays on Saturday or Sunday (but not all on one day). But of these 7 "combinations" there are 8! permutations we can arrange the students in. So this sample space is $7 \cdot 8!$ (by my reasoning)

So the probability is $\frac{7 \cdot 8!}{7^8}$

Now if this reasoning is correct (which I think it is), then I use it to solve the question:

So there are 7 ways to distribute the players to satisfy the condition that they all have their birthday on any two days but not all on one day. For this question it asks for any two days, so the number of ways to choose 2 days from the 7 days of the week is simply: $C_{7,2}$. Of these combinations, there are 7 ways to allocate the players (7 in Day X, 1 in Day Y then 6 in Day X, 2 in Day Y, etc.). So the total number of possibilities is: $C_{7,2} \cdot 7 \cdot 8!$

However I know that I am wrong since this number is greater than the sample space... but I don't understand what I'm doing wrong, I think my initial reasoning (the "Consider") part is off. Can someone please help me?

I do not think your reasoning is correct. Suppose the two days are fixed on Saturday and Sunday, what we are doing is grouping 8 students into two groups such that no group has zero person. This amount will be $2^8-2$ and not $7\times8!$. So overall the probability should be $\frac{{7\choose2}(2^8-2)}{7^8}$.

In your reasoning, consider the following two permutations:

$1,2,3,4,5,6,7,8$

$8,7,6,5,4,3,2,1$

They will generate duplicated results based on your method of partitioning.(with the first sentence and last sentence re-order each other)

• But in our total sample space, we also have duplicates , so with permutations, I am also generating duplicates to get the total probability – q.Then Oct 1 '15 at 8:03
• In the total sample space $7^8$ there are no duplicates. Can you provide an example of why you think there are duplicates? – cr001 Oct 1 '15 at 8:06
• If you do not understand anything let me know. There is no permutation involved in the sample space $7^8$, all we are doing is for each student, there are 7 possible birthdays. – cr001 Oct 1 '15 at 8:33
• Thanks so much! I understand why there are no duplicates in the sample space (I thought there was because the generic formula of $X^9$ implies duplicates in a word-formation problem but in a different context). It's a bit hard to catch that there are $2^8$ combinations of allocating students within 2 groups (i think you did $-2$ to account for the two cases where there are 8 in one or the other day) – q.Then Oct 1 '15 at 15:09
• So for the first question (fixed to Saturday and Sunday), it would just be $2^8$ over the sample space and the second option where we can choose the days is the combination of 7 choose 2 * 2^8(since we have that many ways of arranging them in each combination of days chosen)? – q.Then Oct 1 '15 at 15:10

croo1's answer is correct, I'll just try to reword it a bit.

1. You don't care which 2 days the guys have birthdays on; so you select 2 out 7, order unimportant: $\binom{7}{2}$. This takes care of the last bit of your question.

2. Imagine you have 8 empty slots. There's only one restriction: in either day there must be at least 1 celebrant. I'm sure you are already familiar with Binomial theorem/formula, so it's clear that the total number of birthdays for these 2 days will be $2^8-2$.

3. Since that last value is for EACH selection of 2 days, you need to multiply them: $\binom{7}{2} (2^8-2)$ to get the total number of ways of 8 ppl having birthdays on exactly 2 days. of the week.

Now you can do the rest

• I know the binomial theorem, but how does it tie (or applicable) into this? (.i.e why is the maximum possible ways to divide 8 students into 2 groups given that day x and y must have at least 1 student) = $2^8 - 2$ can be found from the binomial theorem? – q.Then Oct 1 '15 at 18:20
• you can get $2^8$ from the BT by setting $x=1$. The $-2$ part comes from the fact that there must be at least 1 person in each day – Alex Oct 1 '15 at 18:50