A pizzeria sells small, medium, and large pizzas, and offers five diff

Question

A pizzeria sells small, medium, and large pizzas, and offers five different toppings. Its "Three for Thursday" delivery special allows diners to order three one-topping pizzas for the price of one, but with one restriction: all three pizzas must be the same size, and the topping for each pizza must either be the same for all three pizzas or different for each pizza. How many unique orders are available with this special?

A. 25
B. 35
C. 45
D. 55
E. 75

A. 25
B. 35
C. 45
D. 55
E. 75

broall · Answer

To select the size for 3 pizzas, there are $3$ different sizes.

Now, to select the topping for each pizza, if all 3 topping are the same, there are $5$ different selections. 
If each selected topping are different, there are $5C3=10$ different selections.
Hence, there are total $15$ different selections.

Now, the number of unique available orders is $3 \times 15 = 45$. The answer is C.

GMATinsight · Answer

Total Sizes possible = 3
Total Toppings = 5

Case 1 : Same size and same toppings = 3*5 = 15
 
  Case 2 : Same size and Different toppings = 3*5C3 = 3*10 = 30

Total Orders = 15+30 = 45

5C3 is the total ways to select 3 toppings out of 5 available toppings

Answer: Option C

Kurtosis · Answer

Sizes of pizza available: Small, Medium, Large

Number of different topings available: 5

Condition for orders eligible for 'Three for Thursday' --> Either three pizzas having same topings or three pizzas having different topings

For one particular sized pizza we can have: 5C1 (same topping) + 5C3 (different topping) = 5 + 10 = 15 different orders

We have 3 sizes of pizzas, so number of unique orders we can have is 15 * 3 = 45

Answer: C

Heseraj · Answer

we need to establish two spots for two cases, one for pizza and one for toppings:

Choose pizza * choose same toppings: 3C1 * 5C1=  3*5=15

Choose Pizza * chose different toppings:3C1 * 5C3=  3*10=30

Total: 15+30=45

Correct Response: Option C

praveenmittal95605 · Answer

number of ways of selecting 3 toppings from 5 = 5C3 = 10 + number of cases when all 3 pizzas are of same type = 5

hence 10+5 = 15

number of size we can have = 3 Hence => 3*15 = 45 ways

ScottTargetTestPrep · Answer

Let’s first determine the number of ways to have different toppings on a particular size (e.g., small) pizza.

Since there are 5 different toppings, the number of ways to select 1 same topping for each of the 3 small pizzas is 5C1 = 5. The number of ways to select 1 different topping for each of the 3 small pizzas is 5C3 = (5 x 4 x 3)/(3 x 2) = 10. Thus, the number of ways to select three 1-topping small pizzas (under the requirement) is 5 + 10 = 15.

Using the same logic, the number of ways to select three 1-topping medium pizzas will be 15 and three 1-topping large pizzas will be 15. Thus the number of ways to select three 1-topping pizzas (under the requirement) is 15 x 3 = 45.

Answer: C

Bunuel · Answer

Please read the discussion above - there are 6 solutions provided.

Hope it helps.

bumpbot · Answer

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A pizzeria sells small, medium, and large pizzas, and offers five diff

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