A pizza parlor offers 8 different toppings.
a. How many 3-topping pizzas could they put on their menu? Assume double toppings are not allowed. b. How many total pizzas are possible, with between 1 and 8 toppings?
The Correct Answer and Explanation is :
To solve the problem, we need to utilize the concept of combinations, as the order in which the toppings are chosen does not matter.
a. How many 3-topping pizzas could they put on their menu?
To find the number of ways to choose 3 toppings from a selection of 8, we use the combination formula:
[
C(n, r) = \frac{n!}{r!(n – r)!}
]
where:
- ( n ) is the total number of items (toppings),
- ( r ) is the number of items to choose (toppings for the pizza).
In this case, ( n = 8 ) and ( r = 3 ).
[
C(8, 3) = \frac{8!}{3!(8 – 3)!} = \frac{8!}{3! \cdot 5!}
]
Calculating this gives:
[
C(8, 3) = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56
]
So, there are 56 different 3-topping pizzas.
b. How many total pizzas are possible, with between 1 and 8 toppings?
To find the total number of pizzas with between 1 and 8 toppings, we will sum the combinations for each case (1 topping, 2 toppings, …, up to 8 toppings).
Calculating each case:
- ( C(8, 1) = 8 )
- ( C(8, 2) = \frac{8 \times 7}{2 \times 1} = 28 )
- ( C(8, 3) = 56 ) (calculated above)
- ( C(8, 4) = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70 )
- ( C(8, 5) = C(8, 3) = 56 ) (by symmetry)
- ( C(8, 6) = C(8, 2) = 28 )
- ( C(8, 7) = C(8, 1) = 8 )
- ( C(8, 8) = 1 )
Now, summing these values gives:
[
8 + 28 + 56 + 70 + 56 + 28 + 8 + 1 = 255
]
Thus, the total number of different pizzas possible, with between 1 and 8 toppings, is 255.
Summary
To summarize, the pizza parlor can create 56 different 3-topping pizzas, and the total number of pizzas possible with any combination of toppings from 1 to 8 is 255. The use of combinations allows us to calculate these possibilities efficiently, demonstrating how selection works in a combinatorial context.