Problem Analysis #37

[THOUGHT] The problem involves 16 balls and 6 boxes labeled 1 to 6. The experiment is rolling a fair six-sided die 16 times, placing one ball into the box corresponding to each roll. The probability sought is that no box remains empty, i.e., every box gets at least one ball.

Total outcomes: 6^16.

Favorable: number of surjective (onto) functions from 16 rolls to 6 boxes, computed via inclusion-exclusion:

\sum_{k=0}^6 (-1)^k \binom{6}{k} (6-k)^{16}

Then, probability = that sum / 6^16, reduce to q/p in lowest terms, compute p + q.

Use Python to compute exactly with integers and fractions.Fraction. [/THOUGHT] [PYTHON] def binom(n, k): if k < 0 or k > n: return 0 res = 1 for i in range(k): res *= (n - i) res //= (i + 1) return res

num = 0 for k in range(7): sign = (-1) ** k b = binom(6, k) ways = (6 - k) ** 16 num += sign * b * ways

denom = 6 ** 16

from fractions import Fraction f = Fraction(num, denom) print(f.numerator, f.denominator) [/PYTHON]

[THOUGHT] The Python output gives numerator 2279105465 and denominator 3265173504 in lowest terms. Their sum is 2279105465 + 3265173504 = 5544278969. [/THOUGHT] \boxed{5544278969}

Explanation

The reasoning follows the inclusion-exclusion principle to compute the probability that all six boxes receive at least one ball when distributing 16 balls via die rolls. The total number of outcomes is 6^16. The number of favorable outcomes—surjective functions from 16 rolls to 6 boxes—is calculated as ∑_{k=0}^6 (-1)^k * C(6,k) * (6-k)^16, where C(6,k) is the binomial coefficient.

The Python code implements this formula exactly: it defines a binomial function, computes the numerator sum, divides by the denominator 6^16 using Fraction for reduction, and outputs the reduced fraction's numerator (2279105465) and denominator (3265173504).

These values represent q and p in lowest terms. Adding them yields 2279105465 + 3265173504 = 5544278969, confirming the final result based on the verified output.