Question:
From n distinct integers, two groups of integers are to be selected with $k_1$ integers in the
first group and $k_2$ integers in the second group, where $k_1$ and $k_2$ are fixed and $k_1 + k_2 ≤ n$.
In how many ways can the selection be made such that the smallest integer in the first
group is larger then the largest integer in the second group?
I arranged these number in increasing. $r_1,r_2,…r_n$.
Best Answer
I preassume here that for $i=1,2$ the $k_i$ chosen integers must be distinct.
Since the smallest integer in the first group is demanded to be larger than the largest integer in the second group this boils down to selecting $k_1+k_2$ integers out of $n$ distinct integers.
After selection we place the $k_1$ largest integers in the first group and the other integers in the second group. The mentioned condition makes that the only choice we have.
So the answer is:$$\binom{n}{k_1+k_2}$$