\nWe need to find the number of ordered triples (a, b, c) such that a + b + c = 16, where a, b, and c are integers between 1 and 6 (inclusive).
\nLet’s list the possible combinations (unordered sets of numbers) that sum to 16, where each number is between 1 and 6:
\n– The maximum sum of two dice is 6+6=12. So the minimum value on one die to reach a sum of 16 must be 16 – 12 = 4.
\n– Possible combinations (sorted):
\n – If the lowest value is 4: 4 + ? + ? = 16. The remaining two numbers must sum to 12. The only way to get 12 with two dice is 6 + 6. So, the combination is {4, 6, 6}.
\n – If the lowest value is 5: 5 + ? + ? = 16. The remaining two numbers must sum to 11. Ways to get 11 with two dice are 5 + 6 or 6 + 5. So, the combination is {5, 5, 6}.
\n – If the lowest value is 6: 6 + ? + ? = 16. The remaining two numbers must sum to 10. Ways are 4+6, 5+5, 6+4. The combinations are {6, 4, 6}, {6, 5, 5}, {6, 6, 4}. Note that {6,4,6} is the same combination as {4,6,6}, and {6,5,5} is the same as {5,5,6}. We have already listed the unique combinations: {4, 6, 6} and {5, 5, 6}.<\/p>\nNow, we count the number of distinct permutations (ways) for each combination:
\n– Combination {4, 6, 6}: The possible outcomes when rolling three dice are (4, 6, 6), (6, 4, 6), and (6, 6, 4). There are 3 distinct permutations. (Using the formula for permutations with repetitions: 3! \/ 2! = 3).
\n– Combination {5, 5, 6}: The possible outcomes are (5, 5, 6), (5, 6, 5), and (6, 5, 5). There are 3 distinct permutations. (Using the formula for permutations with repetitions: 3! \/ 2! = 3).<\/p>\n
Total number of ways = 3 (for {4, 6, 6}) + 3 (for {5, 5, 6}) = 6.
\n<\/section>\n