This is an excellent question!
After observing the patterns, we can clearly see that if we’re talking about k, then we need a total of k + 1 numbers, such that first k-1 are arranged in descending order, the kth number should be minimum, and the number at k+1th position will always be greater than the kth position number(as the kth position number is minimum).
Now, the first task is to select k+1 numbers out of n,
and the second task is to place a number except the minimum of these k+1 numbers at the (k+1)th position,
and the third task is to permute the remaining n-(k+1) numbers which were untouched from the beginning. The n-(k+1) numbers will permute in (n-(k+1))! ways.
So, task1 = nC(k+1)
task 2 = k ways(except minimum, we have k choices)
task3 = (n-(k+1))!
So, total permutations = nC(k+1) * k * (n – (k + 1))!
On computing the answer will come out to be option B.