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?

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.
$P(\text{mmtt}) = \frac{^7C_4\cdot \frac{4!}{2!\cdot 2!}\cdot 5^3}{7^7} = .03187$ 
