#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < n; i++)
#define rep2(i, x, n) for (int i = x; i <= n; i++)
#define rep3(i, x, n) for (int i = x; i >= n; i--)
#define each(e, v) for (auto &e : v)
#define pb push_back
#define eb emplace_back
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define sz(x) (int)x.size()
using ll = long long;
using pii = pair<int, int>;
using pil = pair<int, ll>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
const int MOD = 1000000007;
// const int MOD = 998244353;
const int inf = (1 << 30) - 1;
const ll INF = (1LL << 60) - 1;
template <typename T>
bool chmax(T &x, const T &y) {
    return (x < y) ? (x = y, true) : false;
}
template <typename T>
bool chmin(T &x, const T &y) {
    return (x > y) ? (x = y, true) : false;
}

struct io_setup {
    io_setup() {
        ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout << fixed << setprecision(15);
    }
} io_setup;

int main() {
    int A, B, C, N;
    cin >> A >> B >> C >> N;

    double dp1[N + 1][A + 1][B + 1][C + 1], dp2[N + 1][A + 1][B + 1][C + 1], dp3[N + 1][A + 1][B + 1][C + 1];

    rep(i, N + 1) {
        rep(j, A + 1) rep(k, B + 1) rep(l, C + 1) {
            if (i == 0 || j * k * l == 0) {
                dp1[i][j][k][l] = 0;
                dp2[i][j][k][l] = 0;
                dp3[i][j][k][l] = 0;
                continue;
            }
            double S = (j + k + l) * (j + k + l - 1) / 2;
            double a = j * (j - 1) / 2, b = k * (k - 1) / 2, c = l * (l - 1) / 2;
            double x = S - a - b - c;
            dp1[i][j][k][l] = (x * dp1[i - 1][j][k][l] + a * (dp1[i - 1][j - 1][k][l] + 1) + b * dp1[i - 1][j][k - 1][l] + c * dp1[i - 1][j][k][l - 1]) / S;
            dp2[i][j][k][l] = (x * dp2[i - 1][j][k][l] + a * dp2[i - 1][j - 1][k][l] + b * (dp2[i - 1][j][k - 1][l] + 1) + c * dp2[i - 1][j][k][l - 1]) / S;
            dp3[i][j][k][l] = (x * dp3[i - 1][j][k][l] + a * dp3[i - 1][j - 1][k][l] + b * dp3[i - 1][j][k - 1][l] + c * (dp3[i - 1][j][k][l - 1] + 1)) / S;
        }
    }

    cout << dp1[N][A][B][C] << ' ' << dp2[N][A][B][C] << ' ' << dp3[N][A][B][C] << '\n';
}