PYTHON DIRECT solution with explanation (24ms)
\nclass Solution:\n    def numPrimeArrangements(self, n: int) -> int:\n        \n\t\t#First, understand that there is a one-to-one correspondence between\n\t\t#
```\nclass Solution:\n    def numPrimeArrangements(self, n: int) -> int:\n        \n\t\t#First, understand that there is a one-to-one correspondence between\n\t\t#Prime numbers and prime indices\n\t\t\n\t\t#Second, we must concern ourselves with order. The order of the indices matter\n\t\t# [  1 , 2 , 3 ] is not the same as [ 2, 1, 3]\n\t\t\n\t\t#We divide the problem into two parts:\n\t\t#The ordering of the primes and the ordering of the non primes within their respective indices\n\t\t#That is, primes can only be shuffled between existing prime slots\n\t\t#And non primes can only be shuffled between existing non prime slots\n\t\t\n\t\t#Lets say there are 3 valid slots (don\'t worry about prime or not for the moment ) \n\t\t#with three distinct choices. That means there are 3 valid numbers\n\t\t\n\t\t#When zero slots are chosen, we have 3 choices { X, X, X } ; _ _ _\n\t\t#We choose one. One slot is chosen, we have 2 choices left {X, X} ; X _ _\n\t\t#We choose again. Two slots are chosen. We have one choice left { X } ; X X _\n\t\t\n\t\t#Say we started with {A, B ,C }. We could generate\n\t\t# { A ,B ,C } _ _ _\n\t\t\n\t\t# { B ,C } A _ _\n\t\t\n\t\t# { C } A B _\n\t\t# { B  } A C _\n\t\t\n\t\t# {  } A B C\n\t\t# {  } A C B\t\n\t\t\n\t\t#So we started with 3 choices, Then could only choose 2, Then were forced to choose 1\n\t\t#Because each ordering is uniqie, we could have started with any A, B or C\n\t\t#So 3 ways to start * 2 ways after * 1 one left = 3 * 2 * 1\n\t\t\n\t\t#This can be extended for any amount of choices\n\t\t#If we have 10 things. With each chosen slotting, we have one less\n\t\t# 10, 9, 8, 7 , ... , 3 , 2, 1\n\t\t#In other words we have 10 * 9 * 8 * 7 * ... * 3 * 2 * 1 possibiliites of unique slotting\n\t\t#Notice this is 10!\n\t\t\n        #The mod val to reduce the size of the number\n        MOD_VAL = pow( 10, 9 ) + 7;\n        \n        #Find all the primes in n\n        #I use the sieve of Eratosthenes\n        primes = set( self.sieve( n ) );\n\n        #I find the number of prime slots\n        #And the number of non prime slots\n\t\t#Remember, there is a 1 to 1 correspondence between primes and their indices\n\t\t#Which also means there is a 1 to 1 correspondence between non primes and their indices\n        all_primes = len ( primes );\n        all_non = n - all_primes;\n        \n        \n        #There are n! ways of selecting unique slots of primes\n        placement_of_primes = factorial( all_primes );\n        \n        #There are m! ways of selecting unique slots of non primes\n        placement_of_non = factorial( all_non );\n        \n        #There are a * b ways of choosing the possibilities of scenario a_i <-> b_j\n        #For all k possibilities of i and q possiblities of j\n\t\t\n\t\t#Another way of saying we have all the unique ways of organizing the primes\n\t\t#Each unique way of organizing the primes can be ordered with each unique way or \n\t\t#Organizing the non-primes\n\t\t\n\t\t#So all possibilites would be \n\t\t\n\t\t# n!  * m!\n\t\t#All the possiblities of uniquely re arranging prime slots times all the possiblities\n\t\t#Of uniquely re arraning the non prime slots\n        all_possibilities = placement_of_primes * placement_of_non;\n        \n        return all_possibilities % MOD_VAL;\n    \n    \n    def sieve( self, n ):\n        \n        isPrime = [ True for _ in range( n + 1 ) ];\n        \n        isPrime[ 0 ] = False;\n        isPrime[ 1 ] = False;\n        \n        pointer = 0;\n        \n        length = len ( isPrime );\n        \n        while pointer < length:\n            \n            if isPrime[ pointer ] == False:\n                pointer += 1;\n            else:\n                for i in range( pointer + pointer , n + 1, pointer):\n                    isPrime[ i ] = False;\n                    \n                pointer += 1;\n                \n        primes = [];\n        for i in range( length ):\n            if isPrime[ i ]:\n                primes.append( i );\n        return primes;\n```
[]