## 1) Concise abridged problem statement

Given \(N\) married couples, a magician randomly matches each of the \(N\) men to a distinct woman (uniformly among all \(N!\) perfect matchings). Compute the probability that **exactly \(K\)** men are matched with their true wives.  
Output `0` if the probability is zero; otherwise output it as an **irreducible fraction** `A/B`.

Constraints: \(1 \le N \le 100\), \(0 \le K \le N\).

---

## 2) Detailed editorial (solution idea and proof)

### Model as a permutation
Fix an ordering of men and women so that the true matching is:
\[
1 \leftrightarrow 1,\; 2 \leftrightarrow 2,\; \dots,\; N \leftrightarrow N
\]
Any guessed matching is a bijection from men to women, i.e. a permutation \(\pi\) of \(\{1,\dots,N\}\), where man \(i\) is matched to woman \(\pi(i)\).

“Man \(i\) matched to his true wife” means \(\pi(i)=i\), i.e. a **fixed point** of the permutation.

So the problem becomes:

> Pick a uniform random permutation of size \(N\). What is the probability it has exactly \(K\) fixed points?

### Count permutations with exactly \(K\) fixed points
1. Choose which \(K\) positions are fixed points: \(\binom{N}{K}\).
2. On the remaining \(N-K\) positions, we need a permutation with **no fixed points** (a *derangement*). Let \(D_m\) be the number of derangements of \(m\) elements.

Thus the number of permutations with exactly \(K\) fixed points is:
\[
\binom{N}{K}\, D_{N-K}
\]

Total permutations: \(N!\). Therefore:
\[
P = \frac{\binom{N}{K}\, D_{N-K}}{N!}
\]

### Simplify the fraction
Use \(\binom{N}{K} = \dfrac{N!}{K!(N-K)!}\):
\[
P = \frac{ \frac{N!}{K!(N-K)!} \, D_{N-K} }{N!}
= \frac{D_{N-K}}{K!(N-K)!}
\]

So we only need:
- \(m = N-K\)
- \(D_m\)
- denominator \(K!\cdot m!\)

### Computing derangements \(D_m\)
Use the classic recurrence:
- \(D_0 = 1\)
- \(D_1 = 0\)
- \(D_m = (m-1)\bigl(D_{m-1}+D_{m-2}\bigr)\) for \(m \ge 2\)

This is fast for \(m \le 100\) and exact with big integers. Since \(D_m\) and \(k!\cdot m!\) overflow 64-bit integers for \(N\) up to 100, the C++ solution carries a vendored arbitrary-precision `bigint` type (Python would support this natively).

### Output formatting
- If numerator \(D_m = 0\), print `0` (this happens for \(m=1\), i.e. \(K=N-1\)).
- Otherwise reduce the fraction by \(g=\gcd(\text{num},\text{den})\) and print `num/g` / `den/g`.

### Complexity
- Derangement DP up to 100: \(O(N)\)
- Factorials up to 100: \(O(N)\)
- Big integer arithmetic is trivial at this scale.

---

## 3) C++ Solution

```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

// base and base_digits must be consistent
const int base = 1000000000;
const int base_digits = 9;

struct bigint {
    vector<int> z;
    int sign;

    bigint() : sign(1) {}

    bigint(long long v) { *this = v; }

    bigint(const string& s) { read(s); }

    void operator=(const bigint& v) {
        sign = v.sign;
        z = v.z;
    }

    void operator=(long long v) {
        sign = 1;
        if(v < 0) {
            sign = -1, v = -v;
        }
        z.clear();
        for(; v > 0; v = v / base) {
            z.push_back(v % base);
        }
    }

    bigint operator+(const bigint& v) const {
        if(sign == v.sign) {
            bigint res = v;

            for(int i = 0, carry = 0;
                i < (int)max(z.size(), v.z.size()) || carry; ++i) {
                if(i == (int)res.z.size()) {
                    res.z.push_back(0);
                }
                res.z[i] += carry + (i < (int)z.size() ? z[i] : 0);
                carry = res.z[i] >= base;
                if(carry) {
                    res.z[i] -= base;
                }
            }
            return res;
        }
        return *this - (-v);
    }

    bigint operator-(const bigint& v) const {
        if(sign == v.sign) {
            if(abs() >= v.abs()) {
                bigint res = *this;
                for(int i = 0, carry = 0; i < (int)v.z.size() || carry; ++i) {
                    res.z[i] -= carry + (i < (int)v.z.size() ? v.z[i] : 0);
                    carry = res.z[i] < 0;
                    if(carry) {
                        res.z[i] += base;
                    }
                }
                res.trim();
                return res;
            }
            return -(v - *this);
        }
        return *this + (-v);
    }

    void operator*=(int v) {
        if(v < 0) {
            sign = -sign, v = -v;
        }
        for(int i = 0, carry = 0; i < (int)z.size() || carry; ++i) {
            if(i == (int)z.size()) {
                z.push_back(0);
            }
            long long cur = z[i] * (long long)v + carry;
            carry = (int)(cur / base);
            z[i] = (int)(cur % base);
            // asm("divl %%ecx" : "=a"(carry), "=d"(a[i]) : "A"(cur),
            // "c"(base));
        }
        trim();
    }

    bigint operator*(int v) const {
        bigint res = *this;
        res *= v;
        return res;
    }

    friend pair<bigint, bigint> divmod(const bigint& a1, const bigint& b1) {
        int norm = base / (b1.z.back() + 1);
        bigint a = a1.abs() * norm;
        bigint b = b1.abs() * norm;
        bigint q, r;
        q.z.resize(a.z.size());

        for(int i = a.z.size() - 1; i >= 0; i--) {
            r *= base;
            r += a.z[i];
            int s1 = b.z.size() < r.z.size() ? r.z[b.z.size()] : 0;
            int s2 = b.z.size() - 1 < r.z.size() ? r.z[b.z.size() - 1] : 0;
            int d = ((long long)s1 * base + s2) / b.z.back();
            r -= b * d;
            while(r < 0) {
                r += b, --d;
            }
            q.z[i] = d;
        }

        q.sign = a1.sign * b1.sign;
        r.sign = a1.sign;
        q.trim();
        r.trim();
        return make_pair(q, r / norm);
    }

    friend bigint sqrt(const bigint& a1) {
        bigint a = a1;
        while(a.z.empty() || a.z.size() % 2 == 1) {
            a.z.push_back(0);
        }

        int n = a.z.size();

        int firstDigit = (int)sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
        int norm = base / (firstDigit + 1);
        a *= norm;
        a *= norm;
        while(a.z.empty() || a.z.size() % 2 == 1) {
            a.z.push_back(0);
        }

        bigint r = (long long)a.z[n - 1] * base + a.z[n - 2];
        firstDigit = (int)sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
        int q = firstDigit;
        bigint res;

        for(int j = n / 2 - 1; j >= 0; j--) {
            for(;; --q) {
                bigint r1 =
                    (r - (res * 2 * base + q) * q) * base * base +
                    (j > 0 ? (long long)a.z[2 * j - 1] * base + a.z[2 * j - 2]
                           : 0);
                if(r1 >= 0) {
                    r = r1;
                    break;
                }
            }
            res *= base;
            res += q;

            if(j > 0) {
                int d1 =
                    res.z.size() + 2 < r.z.size() ? r.z[res.z.size() + 2] : 0;
                int d2 =
                    res.z.size() + 1 < r.z.size() ? r.z[res.z.size() + 1] : 0;
                int d3 = res.z.size() < r.z.size() ? r.z[res.z.size()] : 0;
                q = ((long long)d1 * base * base + (long long)d2 * base + d3) /
                    (firstDigit * 2);
            }
        }

        res.trim();
        return res / norm;
    }

    bigint operator/(const bigint& v) const { return divmod(*this, v).first; }

    bigint operator%(const bigint& v) const { return divmod(*this, v).second; }

    void operator/=(int v) {
        if(v < 0) {
            sign = -sign, v = -v;
        }
        for(int i = (int)z.size() - 1, rem = 0; i >= 0; --i) {
            long long cur = z[i] + rem * (long long)base;
            z[i] = (int)(cur / v);
            rem = (int)(cur % v);
        }
        trim();
    }

    bigint operator/(int v) const {
        bigint res = *this;
        res /= v;
        return res;
    }

    int operator%(int v) const {
        if(v < 0) {
            v = -v;
        }
        int m = 0;
        for(int i = z.size() - 1; i >= 0; --i) {
            m = (z[i] + m * (long long)base) % v;
        }
        return m * sign;
    }

    void operator+=(const bigint& v) { *this = *this + v; }
    void operator-=(const bigint& v) { *this = *this - v; }
    void operator*=(const bigint& v) { *this = *this * v; }
    void operator/=(const bigint& v) { *this = *this / v; }

    bool operator<(const bigint& v) const {
        if(sign != v.sign) {
            return sign < v.sign;
        }
        if(z.size() != v.z.size()) {
            return z.size() * sign < v.z.size() * v.sign;
        }
        for(int i = z.size() - 1; i >= 0; i--) {
            if(z[i] != v.z[i]) {
                return z[i] * sign < v.z[i] * sign;
            }
        }
        return false;
    }

    bool operator>(const bigint& v) const { return v < *this; }
    bool operator<=(const bigint& v) const { return !(v < *this); }
    bool operator>=(const bigint& v) const { return !(*this < v); }
    bool operator==(const bigint& v) const {
        return !(*this < v) && !(v < *this);
    }
    bool operator!=(const bigint& v) const { return *this < v || v < *this; }

    void trim() {
        while(!z.empty() && z.back() == 0) {
            z.pop_back();
        }
        if(z.empty()) {
            sign = 1;
        }
    }

    bool isZero() const { return z.empty() || (z.size() == 1 && !z[0]); }

    bigint operator-() const {
        bigint res = *this;
        res.sign = -sign;
        return res;
    }

    bigint abs() const {
        bigint res = *this;
        res.sign *= res.sign;
        return res;
    }

    long long longValue() const {
        long long res = 0;
        for(int i = z.size() - 1; i >= 0; i--) {
            res = res * base + z[i];
        }
        return res * sign;
    }

    friend bigint gcd(const bigint& a, const bigint& b) {
        return b.isZero() ? a : gcd(b, a % b);
    }
    friend bigint lcm(const bigint& a, const bigint& b) {
        return a / gcd(a, b) * b;
    }

    void read(const string& s) {
        sign = 1;
        z.clear();
        int pos = 0;
        while(pos < (int)s.size() && (s[pos] == '-' || s[pos] == '+')) {
            if(s[pos] == '-') {
                sign = -sign;
            }
            ++pos;
        }
        for(int i = s.size() - 1; i >= pos; i -= base_digits) {
            int x = 0;
            for(int j = max(pos, i - base_digits + 1); j <= i; j++) {
                x = x * 10 + s[j] - '0';
            }
            z.push_back(x);
        }
        trim();
    }

    friend istream& operator>>(istream& stream, bigint& v) {
        string s;
        stream >> s;
        v.read(s);
        return stream;
    }

    friend ostream& operator<<(ostream& stream, const bigint& v) {
        if(v.sign == -1) {
            stream << '-';
        }
        stream << (v.z.empty() ? 0 : v.z.back());
        for(int i = (int)v.z.size() - 2; i >= 0; --i) {
            stream << setw(base_digits) << setfill('0') << v.z[i];
        }
        return stream;
    }

    static vector<int> convert_base(
        const vector<int>& a, int old_digits, int new_digits
    ) {
        vector<long long> p(max(old_digits, new_digits) + 1);
        p[0] = 1;
        for(int i = 1; i < (int)p.size(); i++) {
            p[i] = p[i - 1] * 10;
        }
        vector<int> res;
        long long cur = 0;
        int cur_digits = 0;
        for(int i = 0; i < (int)a.size(); i++) {
            cur += a[i] * p[cur_digits];
            cur_digits += old_digits;
            while(cur_digits >= new_digits) {
                res.push_back(int(cur % p[new_digits]));
                cur /= p[new_digits];
                cur_digits -= new_digits;
            }
        }
        res.push_back((int)cur);
        while(!res.empty() && res.back() == 0) {
            res.pop_back();
        }
        return res;
    }

    typedef vector<long long> vll;

    static vll karatsubaMultiply(const vll& a, const vll& b) {
        int n = a.size();
        vll res(n + n);
        if(n <= 32) {
            for(int i = 0; i < n; i++) {
                for(int j = 0; j < n; j++) {
                    res[i + j] += a[i] * b[j];
                }
            }
            return res;
        }

        int k = n >> 1;
        vll a1(a.begin(), a.begin() + k);
        vll a2(a.begin() + k, a.end());
        vll b1(b.begin(), b.begin() + k);
        vll b2(b.begin() + k, b.end());

        vll a1b1 = karatsubaMultiply(a1, b1);
        vll a2b2 = karatsubaMultiply(a2, b2);

        for(int i = 0; i < k; i++) {
            a2[i] += a1[i];
        }
        for(int i = 0; i < k; i++) {
            b2[i] += b1[i];
        }

        vll r = karatsubaMultiply(a2, b2);
        for(int i = 0; i < (int)a1b1.size(); i++) {
            r[i] -= a1b1[i];
        }
        for(int i = 0; i < (int)a2b2.size(); i++) {
            r[i] -= a2b2[i];
        }

        for(int i = 0; i < (int)r.size(); i++) {
            res[i + k] += r[i];
        }
        for(int i = 0; i < (int)a1b1.size(); i++) {
            res[i] += a1b1[i];
        }
        for(int i = 0; i < (int)a2b2.size(); i++) {
            res[i + n] += a2b2[i];
        }
        return res;
    }

    bigint operator*(const bigint& v) const {
        vector<int> a6 = convert_base(this->z, base_digits, 6);
        vector<int> b6 = convert_base(v.z, base_digits, 6);
        vll a(a6.begin(), a6.end());
        vll b(b6.begin(), b6.end());
        while(a.size() < b.size()) {
            a.push_back(0);
        }
        while(b.size() < a.size()) {
            b.push_back(0);
        }
        while(a.size() & (a.size() - 1)) {
            a.push_back(0), b.push_back(0);
        }
        vll c = karatsubaMultiply(a, b);
        bigint res;
        res.sign = sign * v.sign;
        for(int i = 0, carry = 0; i < (int)c.size(); i++) {
            long long cur = c[i] + carry;
            res.z.push_back((int)(cur % 1000000));
            carry = (int)(cur / 1000000);
        }
        res.z = convert_base(res.z, 6, base_digits);
        res.trim();
        return res;
    }
};

int k, n;

void read() { cin >> k >> n; }

bigint derangements(int m) {
    if(m == 0) {
        return bigint(1);
    }
    if(m == 1) {
        return bigint(0);
    }

    bigint d_prev2 = 1, d_prev1 = 0;
    for(int i = 2; i <= m; i++) {
        bigint d_next = (d_prev1 + d_prev2) * (i - 1);
        d_prev2 = d_prev1;
        d_prev1 = d_next;
    }

    return d_prev1;
}

bigint factorial(int m) {
    bigint res = 1;
    for(int i = 2; i <= m; i++) {
        res *= i;
    }
    return res;
}

void solve() {
    // Classify each guessed matching by how it permutes the secret pairing.
    // Fixing exactly k of n pairs means choosing k correct positions and
    // deranging the remaining m = n - k (a permutation with no fixed point).
    // The favourable count is C(n, k) * D_m and the total is n!, so the
    // probability simplifies to D_m / (k! * m!).
    //
    // The derangement number is built with the recurrence
    // D_i = (i - 1) * (D_{i-1} + D_{i-2}), D_0 = 1, D_1 = 0. All quantities are
    // computed with the vendored arbitrary-precision bigint since n can reach
    // 100, then the fraction is reduced by its gcd.

    int m = n - k;
    bigint d = derangements(m);

    if(d.isZero()) {
        cout << 0 << '\n';
        return;
    }

    bigint numerator = d;
    bigint denominator = factorial(k) * factorial(m);

    bigint g = gcd(numerator, denominator);
    numerator /= g;
    denominator /= g;

    cout << numerator << '/' << denominator << '\n';
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

---

## 4) Python solution (independent, detailed comments)

```python
from math import gcd

def derangements(m: int) -> int:
    """
    Compute D_m = number of derangements of m items.
    Recurrence:
      D_0 = 1
      D_1 = 0
      D_m = (m-1) * (D_{m-1} + D_{m-2})
    """
    if m == 0:
        return 1
    if m == 1:
        return 0

    d0, d1 = 1, 0  # D_0, D_1
    for i in range(2, m + 1):
        d0, d1 = d1, (i - 1) * (d1 + d0)
    return d1

def factorial(n: int) -> int:
    """Compute n! as an integer (n <= 100 here)."""
    res = 1
    for x in range(2, n + 1):
        res *= x
    return res

def main() -> None:
    # Input: K N
    k, n = map(int, input().split())

    # We need exactly K fixed points, so the remaining m = N-K points are not fixed
    m = n - k

    # Count derangements among those m points
    num = derangements(m)

    # If there are no such permutations, probability is 0
    if num == 0:
        print("0")
        return

    # Probability = D_m / (k! * m!)
    den = factorial(k) * factorial(m)

    # Reduce fraction
    g = gcd(num, den)
    num //= g
    den //= g

    print(f"{num}/{den}")

if __name__ == "__main__":
    main()
```

---

## 5) Compressed editorial

Treat a matching as a random permutation of size \(N\); correct couples are fixed points.  
Number of permutations with exactly \(K\) fixed points:
\[
\binom{N}{K} D_{N-K}
\]
so probability:
\[
\frac{\binom{N}{K} D_{N-K}}{N!}=\frac{D_{N-K}}{K!(N-K)!}.
\]
Compute \(D_m\) by \(D_0=1\), \(D_1=0\), \(D_m=(m-1)(D_{m-1}+D_{m-2})\).  
If numerator is 0 print `0`, else reduce by gcd and print `A/B`.