// Slower C++ solution:
#include <bits/stdc++.h>

#define rep(i,n) for (int i = (0); i < (n); ++i)

#define rrep(i,n) for(int i = 1; i <= (n); ++i)

#define drep(i,n) for(int i = (n)-1; i >= 0; --i)

#define srep(i,s,t) for (int i = s; i < t; ++i)

#define rng(x) (x).begin(),(x).end()

#define rrng(x) (x).rbegin(),(x).rend()

#define limit(x,l,r) max(l,min(x,r))

#define lims(x,l,r) (x = max(l,min(x,r)))

#define isin(x,l,r) ((l) <= (x) && (x) < (r))

#define show(x) cout << #x << " = " << (x) << endl

#define show2(x,y) cout << #x << " = " << (x) << ", " << #y << " = " << (y) << endl

#define show3(x,y,z) cout << #x << " = " << (x) << ", " << #y << " = " << (y) <<  ", " << #z << " = " << (z) << endl

#define showv(v) rep(i,v.size()) printf("%d%c", v[i], i==v.size()-1?'\n':' ')

#define showv2(v) rep(j,v.size()) showv(v[j])

#define showt(t,n) rep(i,n) printf("%d%c", t[i], i==n-1?'\n':' ')

#define showt2(t,r,c) rep(j,r) showt(t[j],c)

#define showvp(p) rep(i,p.size()) printf("%d %d\n", p[i].first, p[i].second);

#define printv(v) rep(i,v.size()) printf("%d\n", v[i])

#define printt(t,n) rep(i,n) printf("%d\n", t[i])

#define incl(v,x) find(rng(v),x)!=v.end()

#define incls(s,c) s.find(c)!=string::npos

#define fi first

#define se second

#define pb push_back

#define eb emplace_back

#define sz(x) (int)(x).size()

#define pcnt __builtin_popcountll

#define bit(n,k) ((n>>k)&1) // nのk bit目

#define bn(x) ((1<<x)-1)

#define dup(x,y) (((x)+(y)-1)/(y))

#define newline puts("")

#define uni(x) x.erase(unique(rng(x)),x.end())

#define SP << " " <<

#define v(T) vector<T>

#define vv(T) v(v(T))

using namespace std;

using ll = long long;

using vi = vector<int>;

using vvi = vector<vi>;

using vl = vector<ll>;

using vs = vector<string>;

using P = pair<int, int>;

using T = tuple<int, int, int>;

using vp = vector<P>;

using vt = vector<T>;

const int mod = 1000000007;

const double EPS = 1e-9;

const int INF = (1<<30)-1;

//const ll INF = (1LL<<62)-1;

#define dame { puts("No"); return 0;}

#define yn {puts("Yes");}else{puts("No");}

//inline int in() { int x; scanf("%d",&x); return x;}

template<class T = int> inline T in() { T x; cin >> x; return (x);}

template<typename T>inline ll suma(const v(T)& a) { ll res(0); for (auto&& x : a) res += x; return res;}

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }

template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }



int main () {

  int n;

  string s;

  cin >> n >> s;

  ll ans = 0;

  for (int i = 0; i <= 999; ++i) {

    string t = to_string(i);

    while (t.size() != 3) {

      t = '0' + t;

    }

    auto it0 = find(rng(s),t[0]);

    auto it1 = find(it0+1,s.end(),t[1]);

    auto it2 = find(it1+1,s.end(),t[2]);

    //show3(it0-s.begin(),it1-s.begin(),it2-s.begin());

    if (it2 != s.end()) ans++;

  }

  cout << ans << endl;

  return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.1202088591862216x):
#include <bits/stdc++.h>

#define rep(i,n) for (int i = (0); i < (n); ++i)

#define rrep(i,n) for(int i = 1; i <= (n); ++i)

#define drep(i,n) for(int i = (n)-1; i >= 0; --i)

#define srep(i,s,t) for (int i = s; i < t; ++i)

#define all(x) (x).begin(),(x).end()

#define rall(x) (x).rbegin(),(x).rend()

#define limit(x,l,r) max(l,min(x,r))

#define lims(x,l,r) (x = max(l,min(x,r)))

#define isin(x,l,r) ((l) <= (x) && (x) < (r))

#define show(x) cout << #x << " = " << (x) << endl

#define show2(x,y) cout << #x << " = " << (x) << ", " << #y << " = " << (y) << endl

#define show3(x,y,z) cout << #x << " = " << (x) << ", " << #y << " = " << (y) <<  ", " << #z << " = " << (z) << endl

#define showv(v) rep(i,v.size()) printf("%d%c", v[i], i==v.size()-1?'\n':' ')

#define showv2(v) rep(j,v.size()) showv(v[j])

#define showt(t,n) rep(i,n) printf("%d%c", t[i], i==n-1?'\n':' ')

#define showt2(t,r,c) rep(j,r) showt(t[j],c)

#define showvp(p) rep(i,p.size()) printf("%d %d\n", p[i].first, p[i].second);

#define printv(v) rep(i,v.size()) printf("%d\n", v[i])

#define printt(t,n) rep(i,n) printf("%d\n", t[i])

#define incl(v,x) (find(all(v),x)!=v.end())

#define incls(s,c) (s.find(c)!=string::npos)

#define fi first

#define se second

#define pb push_back

#define eb emplace_back

#define sz(x) (int)(x).size()

#define pcnt __builtin_popcountll

#define bit(n,k) ((n>>k)&1) // nのk bit目

#define bn(x) ((1<<x)-1)

#define dup(x,y) (((x)+(y)-1)/(y))

#define newline puts("")

#define uni(x) x.erase(unique(all(x)),x.end())

#define SP << " " <<

#define v(T) vector<T>

#define vv(T) v(v(T))

using namespace std;

using ll = long long;

using ld = long double;

using vi = vector<int>;

using vvi = vector<vi>;

using vl = vector<ll>;

using vvl = vector<vl>;

using vs = vector<string>;

using P = pair<int, int>;

using T = tuple<int, int, int>;

using vp = vector<P>;

using vt = vector<T>;

const int mod = 1000000007;

const double EPS = 1e-9;

//const long double EPS = 1e-14;

const int INF = (1<<30)-1;

const ll LINF = (1LL<<62)-1;

#define dame { puts("No"); return 0;}

#define yn {puts("Yes");}else{puts("No");}

inline int in() { int x; cin >> x; return x;}

inline ll lin() { ll x; cin >> x; return x;}

inline char chin() { char x; cin >> x; return x;}

inline string stin() { string x; cin >> x; return x;}

inline double din() { double x; cin >> x; return x;}

//template<class T = int> inline T in() { T x; cin >> x; return (x);}

template<typename T>inline ll suma(const v(T)& a) { ll res(0); for (auto&& x : a) res += x; return res;}

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }

template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

char itoa(int n) { return n + '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;}



int main () {

  int n;

  string s;

  cin >> n >> s;

  int ans = 0;

  for (int id = 0; id <= 999; ++id) {

    string t;

    int tmp = id;

    rep(j,3) {

      t += '0'+tmp%10;

      tmp /= 10;

    }

    //reverse(all(t));

    int ti = 0;

    rep(i,n) {

      if (s[i] == t[ti]) ti++;

      if (ti == 3) break;

    }

    if (ti == 3) ans++;

  }

  cout << ans << endl;

  return 0;

}