Consider $n$ distinct real numbers: $a_{1}, a_{2}, \ldots, a_{n},$ A permutation $([1],[2], \ldots,[n])$ of the indices $\{1,2, \ldots, n\}$ is said to $V$ -shaped if there exists an integer $r \quad(1 \leq r \leq n)$ such that $a_{[1]} \geq a_{[2]} \geq \ldots \geq a_{[r-1]} \geq a_{[r]} \leq a_{[r+1]} \leq \ldots \leq a_{[n]} .$ Find the total of such $\mathrm{V}$ shaped permutations.
My Approach, As the given n numbers are distinct, so $a_{[r]}$ would be the lowest number So the lowest number can take 1,2,....n positions If it takes the i th position then it has (i-1) items to left. So we need to choose (i-1) items out of n-1 items. So $^{n-1}C_{r-1}$ now as all items are distinct no permutation among these choosen items.
$Am$ $I$ $correct?$
Yes, you are correct. For completion i will put the result then. You end up with $$\sum _{r=1}^n\binom{n-1}{r-1}=2^{n-1},$$ so you can think of this by selecting a subset of $\{a_1,\cdots ,a_n\}$ excluding the minimal element. So there are $n-1$ choices and so $2^{n-1}$ choices to choose the set that is gonna decrease to the minimal element. The remaining elements you put them in ascending order.