"Now that’s combinations without repetition for the selection, but with permutations for the picking order," Enzo explained.
Enzo clapped. "A combinatorial probability with two stages!" Calcolo combinatorio e probabilita -Italian Edi...
Enzo’s eyes sparkled. "Now that is combinatorics with constraints ." "Now that is combinatorics with constraints
First person: 10 choices. Second: 9 choices (different from first). Third: 8 choices (different from first two). [ 10 \times 9 \times 8 = 720 ] [ 10 \times 9 \times 8 = 720
Thus, overall probability that a pizza is made the customers are from three different towns: [ \frac{9}{10} \times \frac{25}{57} = \frac{225}{570} = \frac{45}{114} = \frac{15}{38} \approx 0.3947 ] The Revelation Chiara finished her wine. "Enzo, your pizza game is a lesson in combinatorics and probability."
Choose 1 from town A: 5 ways, 1 from B: 5, 1 from C: 5, 1 from D: 5, but we need exactly 3 towns — so first choose which 3 towns out of 4: (\binom{4}{3} = 4) ways. For each set of 3 towns: choose 1 person from each: (5 \times 5 \times 5 = 125) combinations. Then arrange them in order: (3! = 6) ways. Total favorable ordered selections: [ 4 \times 125 \times 6 = 3000 ]
"But wait!" Luca interrupted. "What if you also require that the three chosen customers are all from different towns, and there are 4 towns with 5 customers each? And the selection without replacement must include one from each town — then what's the probability that a random ordered selection of 3 customers satisfies that?"
