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This is a combinatorics problem where Johnny needs to choose **3 different flavors** from a total of **10 flavors**, and the **order of selection does not matter**. This means the number of ways to pick the flavors can be calculated using combinations. 

### Formula for combinations:
\[
C(n, r) = \frac{n!}{r!(n-r)!}
\]

Where:
- \( n = 10 \) (total number of flavors)
- \( r = 3 \) (number of scoops)

Substituting the values:

\[
C(10, 3) = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120
\]

### Final Answer:
**C. 120**

Would you like me to explain further or explore variations of this question?  

### Related Questions:
1. What happens if Johnny wants to choose 3 scoops but some flavors can be repeated?
2. How would the solution change if the order of selection mattered (permutations)?
3. How many ways can Johnny choose 4 flavors instead of 3 from the same 10 options?
4. If Johnny had 12 flavors to pick from instead of 10, how many ways could he choose 3 flavors?
5. What is the general difference between combinations and permutations?

### Tip:
When solving problems involving selections where the order doesn’t matter, always check if you need to use combinations instead of permutations!

At the ice cream shop, Johnny has 10 different flavors to pick from for his ice cream cone. He gets 3 scoops of different flavors. If the order he picks them in doesn't matter, how many ways can he pick his 3 different flavors for his ice cream cone?

Explore how to calculate the number of ways Johnny can pick 3 different ice cream flavors from 10 available options using combinations. This problem is ideal for students in Grades 10-12 and involves key combinatorics concepts.

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