#include <bits/stdc++.h>
using namespace std;

// 型定義
typedef long long ll;
typedef pair<ll, ll> P;

// forループ
#define REP(i,n) for(ll i=0; i<(ll)(n); ++i)

// 定数宣言
const int INF = 1e9;
const int MOD = 1e9+7;
const ll LINF = 1e18;

// グラフ表現
using Graph = vector<vector<int>>;

// グラフの辺表現
using Edge = map<pair<int,int>,int>;

// n次元配列の初期化。第２引数の型のサイズごとに初期化していく。
template<typename A, size_t N, typename T>
void Fill(A (&array)[N], const T &val){
    std::fill( (T*)array, (T*)(array+N), val );
}

// 最大公約数
ll gcd(ll a,ll b){
   if (a%b == 0) return(b);
   else return(gcd(b, a%b));
}

// 最小公倍数
ll lcm(ll a, ll b){
    return a*b/gcd(a, b);
}

int main()
{
    cout << fixed << setprecision(15);
    ll N;
    string A, B, C;

    cin >> N;
    cin >> A;
    cin >> B;
    cin >> C;

    ll count = 0;
    REP(i, N){
        if((A[i] != B[i]) && (B[i] != C[i]) && (C[i] != A[i])){
            count += 2;
            continue;
        }
        if((A[i] == B[i]) && (B[i] != C[i]) && (C[i] != A[i])){
            count++;
            continue;
        }
        if((A[i] != B[i]) && (B[i] == C[i]) && (C[i] != A[i])){
            count++;
            continue;
        }
        if((A[i] != B[i]) && (B[i] != C[i]) && (C[i] == A[i])){
            count++;
            continue;
        }
    }

    cout << count << endl;
    return 0;
}
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <cmath>
#include <iomanip>
#include <stack>
#include <queue>
#include <numeric>
#include <map>
#include <unordered_map>
#include <set>
#include <fstream>
#include <chrono>
#include <random>
#include <bitset>
#define rep(i,n) for(int i=0;i<(n);i++)
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define sz(x) ((int)(x).size())
#define pb push_back
using ll = long long;
using namespace std;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
ll gcd(ll a, ll b) {return b?gcd(b,a%b):a;}
ll lcm(ll a, ll b) {return a/gcd(a,b)*b;}

void solve(){
  int N; cin >> N;
  string A, B, C; cin >> A >> B >> C;
  int ans = 0;
  rep(i,N){
    set<char> st;
    st.insert(A[i]);
    st.insert(B[i]);
    st.insert(C[i]);
    ans += st.size()-1;
  }
  cout << ans << endl;
}

int main(){
  solve();
  return 0;
}
