#define MOD_TYPE 2

#pragma region Macros

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

#if 1
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#endif

#if 0
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>
using namespace __gnu_pbds;
using namespace __gnu_cxx;
template <typename T>
using extset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#endif

#if 0
#include <boost/multiprecision/cpp_dec_float.hpp>
#include <boost/multiprecision/cpp_int.hpp>
using Int = boost::multiprecision::cpp_int;
using lld = boost::multiprecision::cpp_dec_float_100;
#endif

using ll = long long int;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pld = pair<ld, ld>;
template <typename T>
using smaller_queue = priority_queue<T, vector<T>, greater<T>>;

constexpr ll MOD = (MOD_TYPE == 1 ? (ll)(1e9 + 7) : 998244353);
constexpr int INF = (int)1e9 + 10;
constexpr ll LINF = (ll)4e18;
constexpr ld PI = acos(-1.0);
constexpr ld EPS = 1e-7;
constexpr int Dx[] = {0, 0, -1, 1, -1, 1, -1, 1, 0};
constexpr int Dy[] = {1, -1, 0, 0, -1, -1, 1, 1, 0};

#define REP(i, m, n) for (ll i = m; i < (ll)(n); ++i)
#define rep(i, n) REP(i, 0, n)
#define REPI(i, m, n) for (int i = m; i < (int)(n); ++i)
#define repi(i, n) REPI(i, 0, n)
#define MP make_pair
#define MT make_tuple
#define YES(n) cout << ((n) ? "YES" : "NO") << "\n"
#define Yes(n) cout << ((n) ? "Yes" : "No") << "\n"
#define possible(n) cout << ((n) ? "possible" : "impossible") << "\n"
#define Possible(n) cout << ((n) ? "Possible" : "Impossible") << "\n"
#define all(v) v.begin(), v.end()
#define NP(v) next_permutation(all(v))
#define dbg(x) cerr << #x << ":" << x << "\n";
#define UNIQUE(v) v.erase(unique(all(v)), v.end())

struct io_init {
  io_init() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << setprecision(30) << setiosflags(ios::fixed);
  };
} io_init;
template <typename T>
inline bool chmin(T &a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
inline bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
inline ll CEIL(ll a, ll b) { return (a + b - 1) / b; }
template <typename A, size_t N, typename T>
inline void Fill(A (&array)[N], const T &val) {
  fill((T *)array, (T *)(array + N), val);
}
template <typename T>
vector<T> compress(vector<T> &v) {
  vector<T> val = v;
  sort(all(val)), val.erase(unique(all(val)), val.end());
  for (auto &&vi : v) vi = lower_bound(all(val), vi) - val.begin();
  return val;
}
template <typename T, typename U>
constexpr istream &operator>>(istream &is, pair<T, U> &p) noexcept {
  is >> p.first >> p.second;
  return is;
}
template <typename T, typename U>
constexpr ostream &operator<<(ostream &os, pair<T, U> &p) noexcept {
  os << p.first << " " << p.second;
  return os;
}
random_device seed_gen;
mt19937_64 engine(seed_gen());

#pragma endregion

// --------------------------------------

array<ld, 3> dp[60][60][60][60];

inline ld C2(ld n) { return n * (n - 1) / 2; }

void add(array<ld, 3> &A, array<ld, 3> B) { rep(i, 3) A[i] += B[i]; }
array<ld, 3> mul(array<ld, 3> A, ld t) {
  rep(i, 3) A[i] *= t;
  return A;
}

array<ld, 3> f(int a, int b, int c, int n) {
  if (n == 0) return {0, 0, 0};
  if (dp[a][b][c][n][0] >= 0) return dp[a][b][c][n];
  array<ld, 3> res = {0, 0, 0};
  if (a >= 2) {
    array<ld, 3> tmp = f(a - 1, b, c, n - 1);
    tmp[0] += 1.0;
    add(res, mul(tmp, C2(a) / C2(a + b + c)));
  }
  if (b >= 2) {
    array<ld, 3> tmp = f(a, b - 1, c, n - 1);
    tmp[1] += 1.0;
    add(res, mul(tmp, C2(b) / C2(a + b + c)));
  }
  if (c >= 2) {
    array<ld, 3> tmp = f(a, b, c - 1, n - 1);
    tmp[2] += 1.0;
    add(res, mul(tmp, C2(c) / C2(a + b + c)));
  }
  add(res,
      mul(f(a, b, c, n - 1), (1.0 - (C2(a) + C2(b) + C2(c)) / C2(a + b + c))));
  return dp[a][b][c][n] = res;
}

void solve() {
  array<ld, 3> tmp = {-1, -1, -1};
  Fill(dp, tmp);
  int a, b, c, n;
  cin >> a >> b >> c >> n;
  auto ans = f(a, b, c, n);
  rep(i, 3) cout << ans[i] << (i == 3 - 1 ? "\n" : " ");
}

int main() { solve(); }