what is the probability that in a random group of seven people two were born on Monday and two on Sunday? If people can be born with the same probability any day of the week, what is the probability that in a random group of seven people two were born on Monday and two on Sunday?
My analysis: 
7 people can be born in 7^7 ways.
7 people can be shuffled in 7! Ways.
4 people occupy two days and the 3 remaining people occupy the 5 other days in 5^3
My analysis lead to the following solution: 7!*5^3/7^7
Is that correct?
 A: There are only $^7C_4\cdot \frac{4!}{2!\cdot 2!}\cdot 5^3$ ways to have four out of seven people with two birthdays on Monday and two on Tuesday.
$P(\text{mmtt}) = \frac{^7C_4\cdot \frac{4!}{2!\cdot 2!}\cdot 5^3}{7^7} = .03187$ 
A: First we choose 2 people among 7 that are born on monday and then we choose 2 people among 5 that are born on Sunday. So the number of good outcomes is $${7\choose 2}\cdot {5\choose 2} \cdot 5^3= 7\cdot 3\cdot 5\cdot 2\cdot 5^3 $$
so the probability you seek is $$P = {30\cdot 5^3\over 7^6}$$
A: I will try to make an explanation for Phil:s answer.


*

*We first choose the four people to have birthday on either sunday or monday. This can be done with something called "combinations" sometimes written nCr or $^nC_r$. 4 among 7 can be chosen in $^7C_4$ ways. 

*Then second factor is also an nCr, it is when whe calculate how many ways we can split those 4 people into groups of 2. One group for sunday and one for monday.

*Last we have 3 people left and they can be put in any of the remaining free 5 days and that's where the $5^3=125$ comes from.

