Calcolo Combinatorio E Probabilita -italian Edi... Apr 2026

First person: 10 choices. Second: 9 choices (different from first). Third: 8 choices (different from first two). [ 10 \times 9 \times 8 = 720 ]

Total possible ordered selections (without replacement from 20): (20 \times 19 \times 18 = 6840).

Enzo’s eyes sparkled. "Now that is combinatorics with constraints ."

Enzo clapped. "A combinatorial probability with two stages!" Calcolo combinatorio e probabilita -Italian Edi...

10 possible choices (all mushrooms, all onions, etc.) [ \frac{10}{1000} = \frac{1}{100} ]

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 ]

Enzo nodded. "It happened once. A trio of truffle enthusiasts. The pizza was… intense." A burly farmer named Marco asked, "What about the chance that all three toppings are different?" First person: 10 choices

The catch? The three chosen customers would pick , and the same topping could be chosen more than once. Enzo would then combine their choices into one bizarre, three-topping pizza. The First Mystery One rainy evening, a young data scientist named Chiara sat at the counter.

Enzo smiled, sliding her a free bruschetta . "Ah, combinatoria . Let’s reason."

Each of 3 people chooses 1 topping from 10: [ 10 \times 10 \times 10 = 1000 ] [ 10 \times 9 \times 8 = 720

"So most of the time," Marco laughed, "the pizza is a mix of three distinct flavors!" That night, a boy named Luca asked the most curious question: "What if you drew the names without replacement from a total of 20 customers, but then the three chosen still pick toppings with repetition? And also, before picking toppings, you shuffle a deck of 40 Scoppia cards (Italian regional cards: four suits, numbered 1 to 10). If the first card is a '1' of any suit, you cancel the pizza game. If not, you proceed. What’s the chance we actually make a pizza?"

"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?"

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.)

Total cards: 40. Cards with value 1: 4 (one per suit). [ P(\text{not drawing a '1'}) = \frac{36}{40} = \frac{9}{10} ]

This is always possible once we reach this stage. So the probability that a pizza gets made is just the probability of not drawing a '1' first: