rachelmaddie
Full Member
- Joined
- Aug 30, 2019
- Messages
- 851
Hi. I need this checked please.

6)
Each of the 7 appetizers can be paired with one of the 10 entrees, and each entree can be paired with one of the 6 desserts.
P(A)
The possible outcome when a customer chooses appetizer: 7
P(E)
The possible outcome when a customer chooses entree: 10
P(D)
The possible outcome when a customer chooses dessert: 6
So the total different lunch specials is:
P(A) * P(E) * P(D)
Therefore, the number of combinations is: 7 * 10 * 6 = 420 different lunch specials
6)
Each of the 7 appetizers can be paired with one of the 10 entrees, and each entree can be paired with one of the 6 desserts.
P(A)
The possible outcome when a customer chooses appetizer: 7
P(E)
The possible outcome when a customer chooses entree: 10
P(D)
The possible outcome when a customer chooses dessert: 6
So the total different lunch specials is:
P(A) * P(E) * P(D)
Therefore, the number of combinations is: 7 * 10 * 6 = 420 different lunch specials
