Let $M,F$ be the amount of males and females, and $D=365$ the amount of available days.
Then the total number of configurations that don't have a male-female pair with the same birthday is
$$ \sum_{k=1}^M {D \choose k} k! \, S_{M,k} (D-k)^F $$
Here $k$ is be the amount of different birthdays for the male population; and $S_{M,k}$ is the Stirling number of the second kind.
Hence
$$ p=1- \frac{\sum_{k=1}^M {D \choose k} k! \, S_{M,k} (D-k)^F}{D^{F+M}} $$
a) We solve this problem by first finding the probability of NONE of the $17$ people having the same birthday.
Person $1$ has a birthday (let's call it $D_1$). Then Person $2$ must have a birthday other than $D_1$ (let's call it $D_2$), which leaves him $365-1 = 364$ options. Person $3$ must have a birthday other than $D_1$ and $D_2$ (let's call it $D_3$), which leaves him $365-2 = 363$ options. This goes on until you reach $D_{17}$.
Because there are $365^{17}$ arrangements of birthday possible, and you have $365 \cdot 364 \cdot 363 \cdots 349$ options, the answer is $\displaystyle \frac{365 \cdot 364 \cdot 363 \cdots 349}{365^{17}}$.
Note that we're not finished yet, however, because we found the probability that NONE of them have the same birthday when we really want the probability that AT LEAST TWO have the same birthday. Fortunately, the math is simple - we just subtract the value we found from $1$. So the final answer is $$ 1 - \frac{365 \cdot 364 \cdot 363 \cdots 349}{365^{17}}.$$
b) We can choose two people among the $17$ to have the birthday of January $1$st, and we don't care when the others have their birthdays as long as it's not January $1$st.
We have to choose the two people. Because order doesn't matter, this will be a combination - namely, $\dbinom{17}{2}$. The probability that both of them have the birthday of January $1$st is $\displaystyle \left(\frac{1}{365}\right)^2$.
We then have to assign a birthday to each of the others (N.B.: They can have the same birthdays), which means the probability will be $\displaystyle \left(\frac{364}{365}\right)^{15}$).
Multiplying all of these together, we get
$$\dbinom{17}{2} \left(\frac{1}{365}\right)^2 \left(\frac{364}{365}\right)^{15}.$$
Best Answer
Hint: your approach is a good one. What is the chance if there are only two people? Three?