Question: Let $n \geq 1$ be an integer. Consider a uniformly random permutation of the set $\{1,2,3, \ldots, 2n\}$. Define the event:
A = "both the first element and the last element in the permutation are even integers"
What is $\Pr(A)$?
Answer: $\dfrac{(n-1)}{2(2n-1)}$
Attempt: I'm not sure how to go about this. I know there are $2$ positions out of $n$ positions so $nC2$ and there is $1$ way to have first and last element be even integers. And the $2n-2$ remaining integers we can arrange in $(2n-2)!$ ways. And this is out of $n!$ ways.
So, $\Pr = \dfrac{(nC2) \cdot (2N-2)!}{n!}$
This approach doesn't give me the right answer.
Best Answer
There are $2n$ ways to choose the first number and $2n - 1$ ways to choose the last number. Of these, there are $n$ favorable choices for the first number since it must be even, and $n - 1$ favorable choices for the last number since it must be one of the remaining even numbers. Thus, the probability that both the first and last numbers are even is $$\Pr(A) = \frac{n(n - 1)}{2n(2n - 1)} = \frac{n - 1}{2(2n - 1)}$$
Why is your answer incorrect?
You selected two even numbers to place in the first and last positions, then asserted that there is one way to have the first and last elements be even integers. However, there are two ways to arrange the selected integers in those positions.
You also asserted that there are $n!$ ways to arrange the $2n$ integers in the set $\{1, 2, 3, \ldots, 2n\}$. However, there are $(2n)!$ such arrangements.
With these corrections, you would have obtained $$\Pr(A) = \frac{\binom{n}{2} \cdot 2!(2n - 2)!}{(2n)!} = \frac{\frac{n!}{2!(n - 2)!} \cdot 2!(2n - 2)!}{(2n)!} = \frac{n(n - 1)}{2n(2n - 1)} = \frac{n - 1}{2(2n - 1)}$$