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.

Usernamevisible · Answer

Case 1: All same topping
Choose the size:
3
Choose the topping:
5c1
Total:
3 × 5 = 15
----------

Case 2: All different toppings
 Wrong approach: 
3 × 5 × 4 × 3
 Let's interpret that:
• 3 choices for size
• 5 choices for first topping
• 4 choices for second topping
• 3 choices for third topping
Giving:
3 × 5 × 4 × 3 = 180
The problem is that this treats the pizzas as:
• Pizza #1
• Pizza #2
• Pizza #3
But the problem never labels the pizzas. 
----------------------------------------

Example
Suppose the toppings are:
• Pepperoni
• Mushroom
• Olive
Your method counts all of these separately:

1. P, M, O
2. P, O, M
3. M, P, O
4. M, O, P
5. O, P, M
6. O, M, P
   But these are actually the same order:
 Three pizzas with Pepperoni, Mushroom, and Olive.
 So every valid order is counted:
 3! = 6
 times.

Thus answer for case 2 will be 
3*5c3 = 30

Giving 30+15 = 45 option C

---

Fix your approach
Take your count and divide by 6:
(3 × 5 × 4 × 3) / 3!
= 180 / 6
= 30
So:
Different toppings = 30
Same toppings = 15
Total:
15 + 30 = 45
Answer = 45
-----------

GMAT takeaway
If you naturally write:
5 × 4 × 3
ask yourself:
"Did I just create a first, second, and third slot?"
If the objects are not labeled, divide by the number of ways you can rearrange them:
3! = 6
That's exactly how to repair your approach.

JM007 · Answer

In case 2, why are we selecting the toppings rather than arranging them as 5*4*3?

egmat · Answer

Hi JM007,

Reaching for 5 × 4 × 3 is natural. You're picturing three slots - a first topping, a second, a third - and filling them one by one. That's the right instinct  when the three things you're filling are genuinely different from each other.  Here they aren't, and that's the whole catch.

In Case 2 you order three pizzas of the  same size , each with a different topping. Once they show up, you just have three same-sized pizzas in a box. There's no "first" pizza or "second" pizza - nothing distinguishes them except their toppings. So the order {Pepperoni, Mushroom, Olive} is  one  order, no matter what sequence you list the toppings in.

But 5 × 4 × 3 counts  sequences . It treats P-M-O, M-O-P, O-P-M, ... as different orders. There are 3! = 6 ways to rearrange the same three toppings, so 5 × 4 × 3 = 60 counts every real order  6 times over . Divide that out: 60 ÷ 6 = 10 = 5C3. That's why we  select  (5C3) instead of  arrange .

So when would 5 × 4 × 3 actually be correct?  Only when the three pizzas are genuinely distinguishable - when each topping is attached to a specific, labeled pizza. The question would need to make the pizzas different from one another, for example:

Now small, medium, and large are three distinct slots. Pepperoni on the small vs. pepperoni on the large are genuinely different orders, so you  do  arrange: 5 × 4 × 3 = 60 for the different-topping case.

One-line takeaway:  identical pizzas → select (5C3); labeled / different pizzas → arrange (5 × 4 × 3).  The original question deliberately keeps all three the same size - that's exactly what makes it a selection, not an arrangement.

Answer: C

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

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

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