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The image presents a table of outcomes from rolling two dice, with a question asking how many possible sums fall between 5 and 12 inclusive.

To solve this, we need to count how many of these pairs from the table give a sum in the range from 5 to 12.

Here’s the breakdown of the possible sums:

- **Sum = 5:** (1,4), (2,3), (3,2), (4,1) → 4 outcomes
- **Sum = 6:** (1,5), (2,4), (3,3), (4,2), (5,1) → 5 outcomes
- **Sum = 7:** (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 outcomes
- **Sum = 8:** (2,6), (3,5), (4,4), (5,3), (6,2) → 5 outcomes
- **Sum = 9:** (3,6), (4,5), (5,4), (6,3) → 4 outcomes
- **Sum = 10:** (4,6), (5,5), (6,4) → 3 outcomes
- **Sum = 11:** (5,6), (6,5) → 2 outcomes
- **Sum = 12:** (6,6) → 1 outcome

Now, summing up all the outcomes: 
\[
4 + 5 + 6 + 5 + 4 + 3 + 2 + 1 = 30
\]

Thus, there are **30** different ways to get sums between 5 and 12 inclusive.

Would you like any more details or have further questions? Here are five related questions you might find interesting:

1. What is the probability of rolling a sum of 7 with two dice?
2. How many total outcomes are possible when rolling two dice?
3. What is the probability of rolling a sum less than 5 with two dice?
4. How would the answer change if we considered sums between 6 and 10 inclusive?
5. How does the distribution of sums change if we roll three dice instead of two?

**Tip:** When rolling two dice, the most probable outcome is a sum of 7, as it has the highest number of combinations.

Of the 36 possible outcomes, determine the number for which the sum (for both dice) is from 5 through 12 inclusive.

Discover how to count the number of outcomes when rolling two dice for sums between 5 and 12 inclusive. Learn about the probability and combinatorics involved in this problem, suitable for students in Grades 6-8. Understand the use of the counting principle to find different combinations leading to the desired sums.

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