a menu has a list of seven appetizers. your party will choose three. which of the following represents the number of possible choices? 7/3!, 7c3,or 7!-3!.
7 appetizers, choose 3.
Clearly the answer is 7 choose 3.
To calculate the number of possible choices when selecting 3 appetizers from a menu of 7, you can use the concept of combinations. The formula for combinations is denoted as nCr or "n choose r", where n represents the total number of items and r represents the number of items to be chosen.
In this case, you need to calculate 7 choose 3, which is written as 7C3.
To calculate 7C3, you can use the formula:
7C3 = 7! / (3! * (7 - 3)!)
Here's how the formula works:
- 7! (7 factorial) represents the product of all positive integers from 1 to 7: 7 x 6 x 5 x 4 x 3 x 2 x 1.
- 3! (3 factorial) represents the product of all positive integers from 1 to 3: 3 x 2 x 1.
- (7 - 3)! is equal to 4! (4 factorial), which represents the product of all positive integers from 1 to 4: 4 x 3 x 2 x 1.
Now let's calculate 7C3:
7C3 = 7! / (3! * (7 - 3)!)
= (7 x 6 x 5 x 4 x 3 x 2 x 1) / ((3 x 2 x 1) * (4 x 3 x 2 x 1))
= (7 x 6 x 5) / (3 x 2 x 1)
= 35
Therefore, the number of possible choices when selecting 3 appetizers from a menu of 7 is 35. Hence, the answer is 7C3.