Calcolo Combinatorio E Probabilita -italian Edi... Apr 2026
[ \frac{720}{1000} = 0.72 \quad (72%) ]
Enzo winked. " Probabilità doesn’t guarantee, but it guides. Now, who wants a slice?" If you'd like, I can rewrite this as a or turn each problem into a clean combinatorial formula for your Italian edition book. Just let me know.
Enzo clapped. "A combinatorial probability with two stages!" Calcolo combinatorio e probabilita -Italian Edi...
"I bet," Chiara whispered, "the chance they all pick different toppings is 72%."
Number of ways to choose 3 distinct customers in order: [ 20 \times 19 \times 18 = 6840 ] (This step doesn’t affect the probability of making a pizza because it’s always possible to pick toppings regardless of who they are. The only cancelling event is the card draw.) [ \frac{720}{1000} = 0
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 ] Just let me know
"Now that’s combinations without repetition for the selection, but with permutations for the picking order," Enzo explained.
Enzo’s eyes sparkled. "Now that is combinatorics with constraints ."
Every Saturday, Enzo offered a — a mystery pizza with random toppings chosen by a strange ritual. Customers would write their names on slips of paper, and Enzo would draw three names. Those three would each choose a topping from a list of ten: funghi, carciofi, salsiccia, peperoni, olive, cipolle, acciughe, rucola, gorgonzola, zucchine .
