Westonci.ca is your trusted source for accurate answers to all your questions. Join our community and start learning today! Explore our Q&A platform to find in-depth answers from a wide range of experts in different fields. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

Peter sells 5 flavors of gelato (Italian ice cream) at his stand. Today, he is selling Hazelnut, Chocolate, Tiramisu, Pistachio, and Strawberry. Allen wants to try three different scoops in his cone. Below is a list of possible combinations. What’s missing?

\begin{tabular}{|ccccc|}
\hline
[tex]$H - C - T$[/tex] & [tex]$H - C - S$[/tex] & [tex]$H - T - S$[/tex] & [tex]$C - T - P$[/tex] & [tex]$C - P - S$[/tex] \\
[tex]$H - C - P$[/tex] & [tex]$?$[/tex] & [tex]$H - P - S$[/tex] & [tex]$?$[/tex] & [tex]$T - P - S$[/tex] \\
\hline
\end{tabular}

A. [tex]$H - T - P$[/tex]

B. [tex]$C - T - S$[/tex]

C. [tex]$S - T - H$[/tex]

D. [tex]$P - T - S$[/tex]


Sagot :

Let's identify the missing combinations of three different scoops of gelato from the following list:

1. Hazelnut (H)
2. Chocolate (C)
3. Tiramisu (T)
4. Pistachio (P)
5. Strawberry (S)

First, we list all possible combinations of choosing 3 scoops out of 5 flavors. There are combinatorial ways to do this. Specifically, the number of combinations is calculated by the combination formula [tex]\( \binom{n}{k} \)[/tex] , where [tex]\( n \)[/tex] is the total number of items and [tex]\( k \)[/tex] is the number of items to choose. Here, [tex]\( n = 5 \)[/tex] and [tex]\( k = 3 \)[/tex]:

[tex]\[ \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10 \][/tex]

So, there are 10 possible combinations. Let's list them:

1. H - C - T
2. H - C - S
3. H - T - S
4. C - T - P
5. C - P - S
6. H - C - P
7. H - P - S
8. T - P - S
9. H - T - P
10. C - T - S

Next, we cross-check each of these combinations with the given list to find the missing ones:

| Given List | Found in Complete List? |
|-------------------------|-------------------------|
| H - C - T | Yes |
| H - C - S | Yes |
| H - T - S | Yes |
| C - T - P | Yes |
| C - P - S | Yes |
| H - C - P | Yes |
| ? (First missing) | No |
| H - P - S | Yes |
| ? (Second missing) | No |
| T - P - S | Yes |

By comparing the lists, we determine that the missing combinations in the table are:

For the first missing:
- H - T - P

For the second missing:
- C - T - S

Hence, the complete list with missing combinations filled in would be:
[tex]\[ \begin{tabular}{|ccccc|} \hline H - C - T & H - C - S & H - T - S & C-T-P & C-P-S \\ H - C - P & H - T - P & H - P - S & C - T - S & T-P-S \\ \hline \end{tabular} \][/tex]

The two missing answers from the given options are:
A. H - T - P
B. C - T - S