Random split of 50 nuggets among four students

A teacher buys $50$ chicken nuggets for four students. Three of them (Stan, Kyle, Kenny) each need at least $6$ nuggets to be happy, and the fourth (Eric) needs at least $18$. The teacher divides the nuggets into four piles, one per student, with every split (every ordered quadruple of non-negative counts summing to $50$) equally likely. You learn that each student received at least $5$ nuggets. The conditional probability that all four are happy can be written as $\dbinom{a}{b}\Big/\dbinom{c}{d}$. What is the minimum possible value of $a+2b+c+2d$?

$62$. Conditioning on at least $5$ each leaves $30$ to distribute freely: $\binom{33}{3}$ splits. Meeting all demands leaves $50-36=14$ free: $\binom{17}{3}$ splits. So the probability is $\binom{17}{3}/\binom{33}{3}$, and choosing $b=d=3$ (not $14$ or $30$) gives $17+6+33+6=62$.