Consider all permutations of the integers $1,2,...,100$. I how many of these permutations will the $25th$ number be the minimum of the first $25$ numbers and the $50th$ number be the minimum of the first $50$ numbers?
I am very weak in these counting problems and combinatorics in general. Any help will be appreciated as I have no idea of where to even begin solving this problem. Thank you.
On
If you choose a random permutation $\pi$, then the smallest element of $\pi_1,\pi_2,\dots,\pi_{25}$ is equally likely to be any of the $\pi_i$ for $1\le i\le 25$, so the probability that $\pi_{25}$ is smallest is $\frac1{25}$. Similarly, $\pi_{50}$ has a $\frac1{50}$ chance of being the smallest of the first $50$ elements. Finally, these two events are independent of each other; this is because a random permutation can be chosen by choosing independently, for each element, how it should rank with respect to the previously chosen elements.
Therefore, the probability of a random permutations satisfying your conditions is $\frac1{25}\cdot \frac1{50}$, so the number of permtuations is $\frac{100!}{25\cdot 50}$.
Setup:
Pick what the first 50 numbers are but don't arrange them yet.
The fiftieth number must be the smallest of those chosen.
From those 49 numbers that were chosen which remain, pick which 25 of them appear in the first 25 spaces.
The 25th number must be the smallest of those chosen in the previous step.
Arrange the rest of the 24 selected numbers for the first section.
Arrange the remaining 24 numbers for the second section.
Arrange the final 50 numbers in the latter half.
Apply multiplication principle and conclude.