Find the probability that the 8th woman to appear is in 17th position. There are 25 people, consisting of 15 women and 10 men that are lined up in a random order. Find the probability that 7 women are in position 1 through 16, and a woman in position 17.
How to go about this problem?
 A: HINT: Choose $1$ woman from $15$ and put her to $17^{th}$ position. Then from the rest of $14$, choose $7$ of them; and from $10$ men choose $9$ of them and permute them in the positions $1-16$. Then, permute the rest of people in the positions $18-25$. 
EDIT: Just to verify my answer with the other two answers, the answer is
$$\frac{\dbinom{15}{1}\dbinom{14}{7}\dbinom{10}{9}16!8!}{25!} \approx 0.028$$
A: The probability that exactly $7$ of the first $16$ positions are filled with women is 
$$\frac{\dbinom{15}{7}\dbinom{10}{9}}{\dbinom{25}{16}}$$
since we must select $7$ of the $15$ women and $9$ of the $10$ men for the first sixteen positions.  That leaves $8$ women and $1$ man.  Therefore, if exactly seven women have been placed in the first $16$ positions, the probability that a woman is then placed in the $17$th position is $8/9$.  Hence, the probability that the eighth women appears in the $17$th position is 
$$\frac{\dbinom{15}{7}\dbinom{10}{9}}{\dbinom{25}{16}} \cdot \frac{8}{9}$$ 
A: It is not necessary to distinguish "by person". Distinguishing "by gender" is enough.
$15$ spots are selected for women out of a total of $25$ spots.
You can think of it as selecting from a basket having exactly $16$ red ball, $1$ blue ball and $8$ yellow balls.
The blue balls correspond with the first $16$ spots, the blue ball with the $17$-th spot and the yellow balls with the last $8$ spots.
By taking without replacement $15$ balls the probability on $7$ red balls, $1$ blue ball and $7$ yellow balls is:
$$\frac{\binom{16}7\binom11\binom{8}7}{\binom{25}{15}}=\frac{\binom{16}7\binom{8}7}{\binom{25}{15}}\approx0.028$$
