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  • + 0 comments

    To solve this, count the total bijections from set M to C, which is simply n! (factorial of n). For more updates or help with social services, visit SRD SASSA Check Status to stay informed and supported.

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    (inventing the weel?)
    Python:

    print(__import__('math').factorial(int(input())) )
    
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    Non-recursive python solution...

    n = int(input())
    
    fact = 1
    while n > 1:
        fact *= n
        n -= 1
    
    print(fact)
    
  • + 0 comments
    def factorial(x):
        if x == 1:
            return 1
        else:
            return x * factorial(x-1)
            
    print(factorial(int(input())))
    
  • + 0 comments
    #include<iostream>
    
    int factorial(int n){
        if(n==1){return n;}
        return factorial(n-1)*n;
    }
    
    int main(){
        int n;
        std::cin>>n;
        std::cout<<factorial(n);
        return 0;
    }