f(n) = ways n people can remain single 
       or pair up.

For n-th person there are two choices:
1) n-th person remains single, so only 1 way so  we recur 
   for remaining i.e f(n - 1)   or you can say 1*f(n-1)
2) n-th person pairs up with any of the 
   remaining n - 1 persons. So apart from the 2 people forming a pair for remaining n-2 persons we We get (n - 1) * f(n - 2) ways

int countFriendsPairings(int n) 
    { 
        if(n<=2) 
            return n;
        return countFriendsPairings(n-1)+ (n-1)*countFriendsPairings(n-2);
    }

int countFriendsPairings(int n) 
    { 
         int m=1e9+7;
         long long int dp[n+1];
         dp[0]=1;
         dp[1]=1;
         for (long long int i = 2; i <= n; i++) 
         {
                dp[i] = ((dp[i - 1]%m) + (((i - 1) * dp[i - 2])%m))%m;
         }
         return dp[n];
    }