To determine the number of ways the sum of two dice can be between 7 and 12 inclusive, let's examine the possible outcomes step by step. The table below represents all possible outcomes when two dice are rolled. Each cell shows the result of rolling the first die (Die 1) with a certain face and the second die (Die 2) with another face.
[tex]\[
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
& & \text{Die 2} & & & & \\
\hline
\text{Die 1} & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
1 & (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\
\hline
2 & (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \\
\hline
3 & (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) \\
\hline
4 & (4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \\
\hline
5 & (5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6) \\
\hline
6 & (6,1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6) \\
\hline
\end{array}
\][/tex]
Now, let's filter all pairs [tex]\((i, j)\)[/tex] where [tex]\( \text{sum}(i, j) \ge 7 \)[/tex] and [tex]\( \text{sum}(i, j) \le 12 \)[/tex].
Pair sums:
- Pairs summing to 7: [tex]\((1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\)[/tex]
- Pairs summing to 8: [tex]\((2,6), (3,5), (4,4), (5,3), (6,2)\)[/tex]
- Pairs summing to 9: [tex]\((3,6), (4,5), (5,4), (6,3)\)[/tex]
- Pairs summing to 10: [tex]\((4,6), (5,5), (6,4)\)[/tex]
- Pairs summing to 11: [tex]\((5,6), (6,5)\)[/tex]
- Pairs summing to 12: [tex]\((6,6)\)[/tex]
Now, we simply count the number of valid pairs for each sum:
- Sum = 7: 6 pairs
- Sum = 8: 5 pairs
- Sum = 9: 4 pairs
- Sum = 10: 3 pairs
- Sum = 11: 2 pairs
- Sum = 12: 1 pair
Adding these counts together:
[tex]\[
6 + 5 + 4 + 3 + 2 + 1 = 21
\][/tex]
Therefore, there are [tex]\( 21 \)[/tex] different ways the sum of two dice can be between 7 and 12 inclusive.
