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?

Question

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?

Answer 1

I think the answer is: $$ \binom{100}{50}\times\binom{49}{25}\times24!\times24!\times50! $$

Answer 2

Let us count the fraction of good permutations. $\frac 1{50}$ have the number in $50^{th}$ the lowest of the first $50$, then $\frac 1{25}$ have the number in $25^{th}$ the lowest of the first $25$. As there are $100!$ permutations, the number of good ones is $$\frac {100!}{50\cdot 25}$$

Answer 3

Step 1 : Select the first 50 numbers. Fix the minimum element. Out of the 49 other elements, select the first 25 numbers. Fix the minimum element of these 25 numbers. Then permute the rest.

Step 2 : Once you select the three sets, the minimum of each of these sets is already fixed. So, for each of these sets, you have to permute the other first 24 numbers (1 - 24) and the next 24 numbers (from 26 - 49) and the other 50 numbers (50 - 100).

You can do that in 24!×24!×50! ways.

Therefore, you can do the whole task in (100C50)×(49C25)×24!×24!×50! ways.