Given K ∈ n *, let F: n * → n * satisfy: for any positive integer greater than k Given that K belongs to n *, let F: n * → n * satisfy that for any positive integer n greater than k, f (n) = n-k (1) Let k = 1, then the value of one of the functions f at n = 1 is? (2) Let k = 4, and when n ≤ 4, 2 ≤ f (n) ≤ 3, then the number of different functions f is? The problem implies that for a positive integer n less than or equal to K, its function value should also be a positive integer. I can't see how it is implied in the problem Question 2: I haven't learned the principle of step-by-step counting. Is there any other method

Given K ∈ n *, let F: n * → n * satisfy: for any positive integer greater than k Given that K belongs to n *, let F: n * → n * satisfy that for any positive integer n greater than k, f (n) = n-k (1) Let k = 1, then the value of one of the functions f at n = 1 is? (2) Let k = 4, and when n ≤ 4, 2 ≤ f (n) ≤ 3, then the number of different functions f is? The problem implies that for a positive integer n less than or equal to K, its function value should also be a positive integer. I can't see how it is implied in the problem Question 2: I haven't learned the principle of step-by-step counting. Is there any other method

f: N * → n * denotes that f is a mapping from a set of positive integers to a set of positive integers
So no matter what the relationship between N and K is, f (n) should be a positive integer
(1) When k = 1, the condition f (n) = n-k is only valid for n > 1, and f (1) can be any positive integer
(2) When n > 4, the function value f (n) = n - 4 is determined by the condition
Only the values of n = 1,2,3,4 can be changed
If 2 ≤ f (n) ≤ 3, f (n) is a positive integer, so f (n) can only be 2 or 3
There are two values for f (1), f (2), f (3) and f (4) respectively. If we count by steps, it is 2 × 2 × 2 = 16 possibilities
Step by step counting is really the easiest way to solve this problem
If you haven't learned the principle of step-by-step counting, you have to enumerate (fortunately, it's not too much)
The possible values of F (1), f (2), f (3) and f (4) are as follows:
2,2,2,2; 2,2,2,3;
2,2,3,2; 2,2,3,3;
2,3,2,2; 2,3,2,3;
2,3,3,2; 2,3,3,3;
3,2,2,2; 3,2,2,3;
3,2,3,2; 3,2,3,3;
3,3,2,2; 3,3,2,3;
3,3,3,2; 3,3,3,3.
There are 16 species
If you have to do it, you can count by one to one correspondence (binary corresponds to 0 to 15 integers), but it is abstract and troublesome
In fact, the principle of step-by-step counting is very easy to understand, it is recommended to master as soon as possible

Given K ∈ n *, let F: n * → n * satisfy that for any positive integer n greater than k, f (n) = n-k
Given K ∈ n +, let F: n + → n + satisfy: for any positive integer n: F (n) = n-k greater than k, please answer and give the reasons: (1) let k = 1, then the value of one of the functions f at n = 1 is______ (2) let k = 4, and when n ≤ 4, 2 ≤ f (n) ≤ 3, then the number of different functions f is________ I can't understand the answer!

Analysis: the question implies that for the positive integer n less than or equal to K, its function value should also be a positive integer, but the corresponding rule depends on the meaning of the question
(1) N = k = 1, the condition given in the question "positive integer n greater than k" is not suitable, but the function value must be a positive integer, so the value of F (1) is a constant (positive integer);
(2) K = 4, and N ≤ 4, which is not suitable for the condition "positive integer n greater than k", so the value of F (n) can be chosen from 2 or 3, and then the number of different functions can be obtained by the principle of multiplication
(1) ∵ n = 1, k = 1 and f (1) is a positive integer
‖ f (1) = a (a is a positive integer)
That is, the function value of F (x) at n = 1 is a (a is a positive integer)
(2) ∵ n ≤ 4, k = 4f (n) is a positive integer and 2 ≤ f (n) ≤ 3
Ψ f (1) = 2 or 3 and f (2) = 2 or 3 and f (3) = 2 or 3 and f (4) = 2 or 3
According to the principle of step-by-step counting, 24 = 16 different functions can be obtained
So the answer is (1) a (a is a positive integer)
（2）16

Given K ∈ n +, let the function f: n + → n + satisfy: for any positive integer n: F (n) = n-k which is greater than k,
(1) N = k = 1, the condition "positive integer n greater than k" is not suitable, but the function value must be a positive integer, so the value of F (1) is a constant (positive integer); is 0?
It's still hard to understand. It's too implicit

Given K ∈ n +, let F: n + → n + satisfy: for any positive integer n: F (n) = n-k greater than k, please answer and give
The reasons are as follows: (2) let k = 4, and when n ≤ 4, 2 ≤ f (n) ≤ 3, then the number of different functions f is________ For the online answers, why 16, I still think 8

Because 2 ≤ f (n) ≤ 3, according to the concept of mapping, we can get: 1, 2, 3, 4 can only correspond to 2 or 3, 1 can correspond to 2, can also correspond to 3, there are two corresponding methods, similarly, 2, 3, 4 have two corresponding methods, according to the principle of multiplication, the number of different functions f is equal to 16
It's 2 * 2 * 2 * 2 = 16