Six people (half are female, half are male) for seven chairs. Problem:
Suppose there are $7$ chairs in a row. There are $6$ people that are going to randomly
sit in the chairs. There are $3$ females and $3$ males. What is the probability that
the first and last chairs have females sitting in them?
Answer:
Let $p$ be the probability we seek. Out of $3$ females, only $2$ can be sitting at the end of the row. I consider the first and last chairs to be at the end of the row.
\begin{align*}
p &= \dfrac{ {3 \choose 2 }  3(2) (4)(3)(2) } { 7(6)(5)(4)(3)(2) } \\
p &= \dfrac{ 3(3)(2) (4)(3)(2) } { 7(6)(5)(4)(3)(2) } \\
p &= \dfrac{ 3(4)(3)(2) } { 7(5)(4)(3)(2) } = \dfrac{ 3(3)(2) } { 7(5)(3)(2) } \\
p &= \dfrac{ 18 } { 35(3)(2) } \\
p &= \dfrac{ 3 } { 35 }
\end{align*}
Am I right?
Here is an updated solution.
Let $p$ be the probability we seek. Out of $3$ females, only $2$ can be sitting at the end of the row. I consider the first and last chairs to be at the end of the row.
\begin{align*}
p &= \dfrac{  3(2) (5)(4)(3)(2) } { 7(6)(5)(4)(3)(2) } \\
p &= \dfrac{  (5)(4)(3)(2) } { 7(5)(4)(3)(2) } \\
p &= \dfrac{1}{7}
\end{align*}
Now is my answer right?

Here is an approach that is closest to your thought process. The three females and the three males are all indistinguishable, so we use combinations instead of permutations. Picking any one sex first gives:
$${7 \choose 3} \cdot {4 \choose 3}$$
total possibilities.
Now if two women are already sitting at the ends, then there is one woman and three men left to fill $5$ seats. Choosing the woman first we have:
$$5 \cdot {4 \choose 3}$$
ways and here we can see another simpler approach: where 3 men sit in 4 seats and the women sit in the remaining 3 seats.
Thus the probability is $\frac{5}{7 \choose 3} = \frac{1}{7}$.
