2018 AIME II Problems/Problem 10
Problem
Find the number of functions from to that satisfy for all in .
Solution 1
Just to visualize solution 1. If we list all possible (x,f(x)), from {1,2,3,4,5} to {1,2,3,4,5} in a specific order, we get 5*5 = 25 different (x,f(x))'s. Namely:
(1,1) (1,2) (1,3) (1,4) (1,5) (2,1) (2,2) (2,3) (2,4) (2,5) (3,1) (3,2) (3,3) (3,4) (3,5) (4,1) (4,2) (4,3) (4,4) (4,5) (5,1) (5,2) (5,3) (5,4) (5,5)
To list them in this specific order makes it a lot easier to solve this problem. We notice, In order to solve this problem at least one pair of (x,x) where x€{1,2,3,4,5} must exist.In this case I rather "go backwards". First fixing 5 pairs (x,x), (the diagonal of our table) and map them to the other fitting pairs (x,f(x)). You can do this in 5!/5! = 1 way. Then fixing 4 pairs (x,x) (The diagonal minus 1) and map them to the other fitting pairs (x,f(x)). You can do this in 4x(5!/4!) = 20 ways. Then fixing 3 pairs (x,x) (The diagonal minus 2) and map them to the other fitting pairs (x,f(x)). You can do this in (5x4x3x6x3)/3!2! + (5x4x3x6x1)/3! = 150 ways. Fixing 2 pairs (x,x) (the diagonal minus 3) and map them to the other fitting pairs (x,f(x)). You can do this in (5x4x6x4x2)/2!3! + (5x4x6x4x2)/2!2! + (5x4x6x2x1)/2!2! = 380 ways. Lastely, fixing 1 pair (x,x) (the diagonal minus 4) and map them to the other fitting pairs (x,f(x)). You can do this in 5!/4! + 4x(5!/3!) + 5! = 205
So 1 + 20 + 150 + 380 + 205 = 756
Solution 2
We can do some caseworks about the special points of functions for . Let , and be three different elements in set . There must be elements such like in set satisfies , and we call the points such like on functions are "Good Points" (Actually its academic name is "fixed-points"). The only thing we need to consider is the "steps" to get "Good Points". Notice that the "steps" must less than because the highest iterations of function is . Now we can classify cases of “Good points” of .
One "step" to "Good Points": Assume that , then we get , and , so .
Two "steps" to "Good Points": Assume that and , then we get , and , so .
Three "steps" to "Good Points": Assume that , and , then we get , and , so .
Divide set into three parts which satisfy these three cases, respectively. Let the first part has elements, the second part has elements and the third part has elements, it is easy to see that . First, there are ways to select for Case 1. Second, we have ways to select for Case 2. After that we map all elements that satisfy Case 2 to Case 1, and the total number of ways of this operation is . Finally, we map all the elements that satisfy Case 3 to Case 2, and the total number of ways of this operation is . As a result, the number of such functions can be represented in an algebraic expression contains , and :
Now it's time to consider about the different values of , and and the total number of functions satisfy these values of , and :
For , and , the number of is
For , and , the number of is
For , and , the number of is
For , and , the number of is
For , and , the number of is
For , and , the number of is
For , and , the number of is
For , and , the number of is
For , and , the number of is
For , and , the number of is
For , and , the number of is
Finally, we get the total number of function , the number is
~Solution by (Frank FYC)
Note (fun fact)
This exact problem showed up earlier on the 2011 Stanford Math Tournament, Advanced Topics Test.
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