Counting Permutation of a set with the condition that $a_1 < a_2$

Question

I'm not sure how the answer is a. How I approached this is:

For $a_1$ < $a_2$ we can select $a_2$ to be $n$ (1 way) and then we have no choices for $a_1$ because it has to be less than $a_2$. Or actually, then it should be $n-1$. Then the remaining ways to arrange it can be done in $(n-2)!$ ways.

So, I get: $n×(n-1)×(n-2)!$ as the total permutations. Now, what I realized it is that what I got and options b and care actually the same thing surprisingly! So with that intuition, option a has to be right. But I really don't get why.

Like if I take a set with {$1,2,3,4$} then I pick $a_2$ to be $4$, my options for $a_1$ will be $1$,$2$ or $3$. But I have it as $n-1$ which will only give me $3$. I also tried $\frac{n-1}{2}$ but that would not give me an integer for some cases. I'm just so confused on how to get an expression that makes sense to me. I

Answer 1

Think about it this way:

There are $n!$ permutations without any restrictions, and by symmetry there are just as many permutations where the first element is smaller than the second, as there are permutations where the first element is greater than the second.

Or: for every permutations with $a_1 < a_2$ you can get the same permutation but with $a_1$ and $a_2$ swapped, and thus one where $a_2 < a_1$, and vice versa. So, for every one where $a_1<a_2$ you have one where $a_2<a_1$, so the ones where $a_1<a_2$ is half of all of them.

Answer 2

The number of all permutation is $n!$. Half of it with $a_1<a_2$ and the other half with $a_2<a_1$.

Answer 3

The condition is only on the first 2, and not on the rest. Once the first two satisfy your requirement the reamining can be arranged in any way, that is $(n-2)!$ Now the first two have are two elements from $n$ with specific order (ascending), and no rearrangement which is ${n\choose 2}$. So it is the product of these two numbers, ${n\choose 2}\times (n-2)!$ which is same as $\frac12 n!$