# 1175. Prime Arrangements

Return the number of permutations of 1 to n so that prime numbers are at prime indices (1-indexed.)

(Recall that an integer is prime if and only if it is greater than 1, and cannot be written as a product of two positive integers both smaller than it.)

Since the answer may be large, return the answer modulo 109 + 7.

Example 1:

Input: n = 5
Output: 12
Explanation: For example [1,2,5,4,3] is a valid permutation, but [5,2,3,4,1] is not because the prime number 5 is at index 1.
Example 2:

Input: n = 100
Output: 682289015

Constraints:

1 <= n <= 100

## Solution

sieve to count primes
care for overflow

O(N) O(N)

## Code

``````class Solution {
private int modNum = 1000000007;
public int numPrimeArrangements(int n) {
int[] mark = new int[n+1];
mark = 1;
mark = 1;
for(int i = 2; i <=n; i++){
for(int j = i; i*j <=n; j++) mark[i*j] = 1;
}
int numOfPrimes = 0;
for(int i = 2; i <= n; i++){
if(mark[i] == 0) numOfPrimes++;
}

long a = factorial(numOfPrimes);
long b = factorial(n - numOfPrimes);

long res = (a * b) % modNum;
return (int)res;

}
long factorial(int num){
long res = 1;
for(int i = 1; i<= num;i++){
res = ((res % modNum) * (i % modNum)) % modNum;
}
return res;
}
}
``````