Three dice, whose all six faces are marked ‘1’ to ‘6’, are thrown. The

Three dice, whose all six faces are marked ‘1’ to ‘6’, are thrown. The number of ways of getting a sum of 16 is

3
4
6
12
This question was previously asked in
UPSC CAPF – 2012
The number of ways of getting a sum of 16 when three dice are thrown is 6.
We 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).
Let’s list the possible combinations (unordered sets of numbers) that sum to 16, where each number is between 1 and 6:
– 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.
– Possible combinations (sorted):
– 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}.
– 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}.
– 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}.

Now, we count the number of distinct permutations (ways) for each combination:
– 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).
– 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).

Total number of ways = 3 (for {4, 6, 6}) + 3 (for {5, 5, 6}) = 6.

Listing out combinations and then calculating permutations is a systematic way to solve such problems. For larger numbers of dice or higher sums, more advanced techniques involving generating functions might be used, but for small cases like this, enumeration is feasible.