combinatorics - In how many permutations of $1,2,3...100$ will the 25th number be the minimum of the first 25 numbers, and likewise for the 50th of the first 50?

Consider all permutations of integers $1,2,3.....100$. In how many of these permutations will the $25^{th} $ number be the minimum of the first 25 numbers and the $50^{th} $ number be the minimum of the first 50 numbers?

My attempt:
Total number of permutations: $100!$

The first 25 numbers can arranged in this manner:
$100,99,98......76,75\Rightarrow \text{the total number of permutations}=24!$

The arrangement of the next 25 numbers:
$74,73,72..............51,50\Rightarrow \text{permutations possible} =24! $

The last 50 numbers can be arranged in $50!$ ways

$\therefore \text{the sequence becomes}: 100,99,98, .....\color{red}{75},74,73, .....\color{red}{50},49,48,......1 $

The total permutations of such sequences are : $ 24!\times24!\times 50! \times 2$

$[\times{ 2}\text{ because the numbers can also be arranged in 50,49,48...25,24...1,51,52...100}]$

I am sure that the answer is not correct. Any ideas?