Like with any math problem, my first objective is to code it out and find the answer. After simplifying the problem a little with some algebra,I wrote the following python code .
import random
# Using a set to store unique tuples
unique_triples = set()
for i in range(1, 100000):
num1 = random.randrange(1, 13)
num2 = random.randrange(1, 14 - num1)
num3 = 14 - num1 - num2
# Create a sorted tuple to avoid permutations of the same numbers
triple = tuple(sorted([num1, num2, num3]))
# Ensure all numbers are unique and the tuple is not already in the set
if len(set(triple)) == 3:
unique_triples.add(triple)
# Convert the set to a list if needed
unique_triples_list = list(unique_triples)
print(len(unique_triples_list))
print(unique_triples_list)
Then, I set out to work out why the answer was what it was with math. Firstly, we need to prove the famous stars and bars and the n choose k function.
Imagine I have n cards, and I want to choose k of them but I do not care about the order in which I pick them. How many ways are there to pick k cards. Well, I have n choices for the first card all the way down to k cards for the kth. This can be written as n!/(n - k)!. However, we overcount each one by k! as we did not consider that we do not care which order we pick them, so the new function is : n!/k!(n - k)!. This is known as the choose function.
Now imagine I have n stars and have k bars, such all bars have at least 1 star in them, and that the bars are labled but the stars are not. Now, instead of thinking of assigning the stars to the bars, think of assigning the bars to the stars by placing partitions between them. Here, we would need k-1 partitions. Also we have n-1 places to put them, as after every star except the last star. In other words I have n-1 valid solutions, and I have to place k-1 of them. This is n-1 choose k-1.
Now it is time to bring the proof together.If the length is five and the mean is 4, the sum of all the numbers must be 20. Also, if the mode is 3 there must be at least 2 instances of the number 3.This brings the total for the other numbers down to 14. If we think of there being 14 stars and the bars being the three numbers, we get the total number of solutions being 78. However, a portion of them contain duplicates of a number which we cant have, as then the mode would not be 3.Also, this includes duplicates of the same sets but in a different order. To remove the sets with multiple numbers, not that there can not be 3 duplicates as 14 is not a multiple of 3 . Think of the first number being n. Then, the folowing numbers are (14 - n)/2.This means n is an even number, as otherwise the duplicates will not be a whole number. There are 6 even numbers under 14. However, there are more sets than with duplicates because of the duplicate set problem. Sorry if the amount of duplicates confuse you! Convince yourself that there are 3 ways of rearranging a set with 2 duplicates. Then, there are 18 sets with a duplicate taking that out from 78, we get 60. As there are n! ways to rearrange a set with no duplicates for length n, this brings the amount of sets down to 10.However we are missing one out : the 3, 3, 8 solution. This was counted out because it had a duplicate, but as three is the duplicate number is 3, this is a valid set. Therefore, the total distinct ways of a valid set is 11, which is what my program said.
Footnote:
Thanks for such an intresting problem : I came up with the proof at night like I usually do. I haven't encountered such a nice problem for a long time, and it is nice to put the thereoms I know to use. Couldn't thank you more!