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?

An alternative approach is that you have the outer seats both occupied with probability $\frac57$ and the conditional probability that they both have a woman is $$\frac{\binom32}{\binom62}=\frac15.$$
So the probability is $\frac57\cdot \frac15=\frac17.$
