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?

• The identical question was asked here – Ross Millikan May 13 '18 at 19:45
• @RossMillikan this question came in an exam today. – DRPR May 13 '18 at 20:16
• I dont know.... – DRPR May 14 '18 at 7:41

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 $$\frac {100!}{50\cdot 25}$$
• @Shailesh: No, whatever $25$ numbers are chosen for the first $25$ there is one smallest. As long as that one is placed in $25^{th}$ position we succeed. Try it with five numbers and asking that the third is smallest of the first three. There are $5!=120$ orders, of which $\frac 13 or 40$ have the third smallest of the first three. – Ross Millikan May 14 '18 at 13:39
I think the answer is: $$\binom{100}{50}\times\binom{49}{25}\times24!\times24!\times50!$$