Detailed mathematical approach

Problem Summary

Let \(12n\) musicians be split on the first day into \(m=3n\) quartets. On the second day they must be split into \(t=4n\) trios, and no pair of musicians is allowed to play together twice. Because every pair inside a first-day quartet has already met, a valid second-day trio can contain at most one musician from each quartet. The task is to count all valid second-day schedules modulo

$10^9+7.$

Mathematical Approach

The fixed first-day partition gives \(m\) distinguishable quartets. The whole problem becomes: how many ways can \(t\) trios be formed so that every trio draws one musician from three different quartets, while each quartet contributes all four of its musicians exactly once?

Step 1: Encode the schedule by an incidence matrix

Introduce an \(m\times t\) matrix \(A\) with entries in \(\{0,1\}\):

$A_{q,r}=1 \iff \text{trio }r\text{ contains one musician from quartet }q.$

The no-repeat condition forces \(A_{q,r}\) to be binary, because a trio cannot take two musicians from the same quartet.

Each trio has exactly three musicians, so every column sum is

$\sum_{q=1}^{m} A_{q,r}=3.$

Each quartet contains four musicians and each of them must appear on day 2 exactly once, so every row sum is

$\sum_{r=1}^{t} A_{q,r}=4.$

Therefore the first combinatorial subproblem is to count \(0\)-\(1\) matrices with row sum \(4\) and column sum \(3\).

Step 2: Track only how many times each quartet has already been used

Process the trio columns from left to right. After some number of processed trios, every quartet has been used \(0\), \(1\), \(2\), \(3\), or \(4\) times. Quartets already used \(4\) times are finished and never matter again, so the state only needs

$\mathbf{c}=(c_0,c_1,c_2,c_3),$

where \(c_j\) is the number of quartets used exactly \(j\) times so far. The number used \(4\) times is implicit:

$c_4=m-c_0-c_1-c_2-c_3.$

Initially no day-2 trio has been built, hence

$\mathbf{c}_{\text{start}}=(m,0,0,0).$

Step 3: Count one transition

To build the next trio, choose

$x_j \text{ quartets from class }j \qquad (j=0,1,2,3),$

subject to

$x_0+x_1+x_2+x_3=3,\qquad 0\le x_j\le c_j.$

This means the trio takes one musician from \(x_0\) previously unused quartets, one musician from \(x_1\) quartets already used once, and so on. The number of ways to make that choice is

$\binom{c_0}{x_0}\binom{c_1}{x_1}\binom{c_2}{x_2}\binom{c_3}{x_3}.$

There are only

$\binom{3+4-1}{4-1}=\binom{6}{3}=20$

possible quadruples \((x_0,x_1,x_2,x_3)\), so each state has a very small transition menu.

Step 4: Update the state and form the recurrence

Every chosen quartet moves from usage class \(j\) to usage class \(j+1\). Therefore

$c_0'=c_0-x_0,$

$c_1'=c_1-x_1+x_0,$

$c_2'=c_2-x_2+x_1,$

$c_3'=c_3-x_3+x_2.$

If \(D_s(c_0,c_1,c_2,c_3)\) denotes the number of ordered partial incidence matrices after \(s\) processed trios, then the recurrence is

$D_{s+1}(c_0',c_1',c_2',c_3') \mathrel{+}= D_s(c_0,c_1,c_2,c_3)\prod_{j=0}^{3}\binom{c_j}{x_j}.$

The base case is

$D_0(m,0,0,0)=1.$

After all \(t\) trio columns have been processed, every quartet must have been used four times, so the terminal state is

$D_t(0,0,0,0).$

This quantity is the number of ordered incidence matrices. Call it \(N_{\text{mat}}\).

Step 5: Convert matrix patterns into actual musician schedules

Once an incidence matrix is fixed, each quartet is attached to exactly four trio columns. Its four labeled musicians can be assigned to those four appearances in

$4!=24$

ways. Different quartets act independently, so the total refinement factor is

$24^m.$

However, the dynamic program processed the trio columns in an artificial order. A finished day-2 schedule is an unordered collection of \(t\) distinct trios, and every such schedule is counted once for each permutation of those \(t\) trio columns. Therefore we divide by \(t!\):

$f(12n)=N_{\text{mat}}\cdot 24^m\cdot \frac{1}{t!}\pmod{10^9+7}.$

Because \(10^9+7\) is prime, \((t!)^{-1}\) is computed with modular exponentiation using Fermat's little theorem.

Worked Example: \(12\) musicians

For \(12\) musicians we have \(n=1\), so

$m=3,\qquad t=4.$

There are exactly three first-day quartets. Every second-day trio must use three different quartets, so each trio must take one musician from each of the three quartets. Hence the incidence matrix is forced: every one of its \(3\times 4\) entries is \(1\). Thus

$N_{\text{mat}}=1.$

Now each quartet distributes its four musicians across the four trio positions in \(4!\) ways, giving

$24^3$

ordered assignments. Since the four trios themselves are unordered, divide by

$4!=24.$

So the final count is

$f(12)=\frac{24^3}{24}=24^2=576,$

which matches the checkpoint used by the implementation.

How the Code Works

The C++, Python, and Java implementations all follow the same plan. They compute \(m=3n\) and \(t=4n\), precompute the binomial values \(\binom{i}{0},\binom{i}{1},\binom{i}{2},\binom{i}{3}\) for \(0\le i\le m\), and enumerate the \(20\) admissible pick patterns \((x_0,x_1,x_2,x_3)\).

The dynamic program stores only the currently reachable states \((c_0,c_1,c_2,c_3)\) in a hash map, which keeps the state space compressed. For each processed trio, the implementation loops over the current states, applies every legal pick pattern, multiplies by the corresponding product of binomial coefficients, and accumulates the result for the next layer modulo \(10^9+7\).

After \(t\) layers, the implementation reads the value at the final state \((0,0,0,0)\), multiplies by \(24^m\), computes \(t!\) modulo \(10^9+7\), inverts that factorial by fast modular exponentiation, and multiplies once more. The small checkpoints \(f(12)=576\) and \(f(24)=509089824\) are used to verify that the recurrence has been implemented correctly.

Complexity Analysis

Let \(S\) be the maximum number of reachable DP states in any layer. Each state tries only \(20\) transition patterns, so the main loop runs in

$O(20tS)=O(tS)$

time. The binomial precomputation costs \(O(m)\), and the modular exponentiations for \(24^m\) and \((t!)^{-1}\) are negligible compared with the state transitions. The memory usage is

$O(S),$

because only the current and next hash-map layers are stored.

Footnotes and References

1. Problem page: https://projecteuler.net/problem=475
2. Incidence matrix: Wikipedia — Incidence matrix
3. Dynamic programming: Wikipedia — Dynamic programming
4. Binomial coefficient: Wikipedia — Binomial coefficient
5. Fermat's little theorem: Wikipedia — Fermat's little theorem