Imagine that you are going to buy ice cream. You will buy 2 balls, and can choose between the flavors strawberry, chocolate, vanilla and blueberry.

a)How many different ice creams can you put together? Show the different possibilities to show how you think.

b)Justify why you are sure that you have included all the possibilities.

c)What if there are more flavors to choose from? How many possibilities are there then?

Let's break down the problem step by step.

### Part (a): Counting the Different Ice Cream Combinations

You are going to buy 2 scoops of ice cream, and you have 4 flavors to choose from: strawberry (S), chocolate (C), vanilla (V), and blueberry (B). The two scoops can be of the same flavor or different flavors. To find the number of possible combinations, we'll look at two cases:

1. **Case 1: Two scoops of the same flavor**
   - You can choose 2 scoops of any of the four flavors. These are the possibilities:
     - (S, S)
     - (C, C)
     - (V, V)
     - (B, B)

   So, there are 4 possibilities in this case.

2. **Case 2: Two scoops of different flavors**
   - If the scoops are of different flavors, you can combine any two of the four flavors. Since order does not matter, we can simply choose pairs of flavors:
     - (S, C)
     - (S, V)
     - (S, B)
     - (C, V)
     - (C, B)
     - (V, B)

   There are 6 possibilities in this case.

Combining the two cases, the total number of different ice cream combinations is \( 4 + 6 = 10 \).

#### Listing all the combinations:
1. (S, S)
2. (C, C)
3. (V, V)
4. (B, B)
5. (S, C)
6. (S, V)
7. (S, B)
8. (C, V)
9. (C, B)
10. (V, B)

Thus, there are **10 different ice creams** you can create.

### Part (b): Justification for Including All Possibilities

We are confident that we've included all possibilities because:
- We considered both cases: scoops of the same flavor and scoops of different flavors.
- We listed all the combinations for both cases and ensured that no combinations were repeated.
  
This systematic approach guarantees that all possible pairings of two scoops from the available flavors have been covered.

### Part (c): Adding More Flavors

Now, let's consider the situation where there are more than 4 flavors. Suppose there are \( n \) flavors. The number of ways to select 2 scoops (considering the possibility of scoops being the same or different) can be calculated using a combination formula.

The number of ways to choose 2 scoops from \( n \) flavors (where order doesn't matter, and scoops can be repeated) is given by the combination formula:

\[
\text{Number of combinations} = \binom{n+1}{2} = \frac{n(n+1)}{2}
\]

This formula accounts for both cases: 
- Selecting two scoops of the same flavor (which gives \( n \) options), and 
- Selecting two different flavors (which gives \( \binom{n}{2} \) options).

Thus, for any number \( n \) of flavors, the total number of possibilities is \( \frac{n(n+1)}{2} \).

### Examples with More Flavors:
- If there are 5 flavors, the number of combinations would be \( \frac{5 \times 6}{2} = 15 \).
- If there are 6 flavors, the number of combinations would be \( \frac{6 \times 7}{2} = 21 \).

Would you like more details or have any questions?

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Here are 5 related questions to expand your understanding:
1. How would the answer change if the order of scoops mattered?
2. What if you could choose up to 3 scoops? How many combinations are possible with 4 flavors?
3. How would you calculate the total number of combinations if you could choose multiple scoops from a larger set of flavors?
4. What if some flavors were not available? How would you adjust the calculation?
5. How would adding constraints (e.g., only 1 chocolate scoop allowed) affect the number of possibilities?

**Tip:** When counting combinations, always consider whether the order of selection matters or not. This changes the formula and the total count!

Explore how to calculate the number of different ice cream combinations when choosing 2 scoops from 4 flavors. Includes a detailed breakdown of cases and formulas used in combinatorial counting. Suitable for Grades 7-9.

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