This is a dynamic programming problem where there is a base case (Base Case: With $0$ slots, there is $1$ way to create $0$ GB of RAM) and induction (Induction: Using the number of ways to create any sum up to the maximum with $n-1$ slots, we can find the number of ways to get a sum to $s$ with $n$ slots by summing the number of ways to get a sum to $s-a$ with $n-1$ slots for all addends $a$). This is also similar to the subset sum, but instead of figuring out whether or not we can get a sum with $1024$ GB, we're finding the number of ways we can do so. This means that this problem is NP-complete, which basically means the time to solve the problem grows very fast as the addends and/or the sum are added with more digits.
Thus, here is my dynamic programming algorithm for this problem in Python:
# These are our addends:
addends = [1, 2, 4, 8, 16, 32, 64]
# num_sums[num_addends][sum] is the number of ways sum can be found
# using num_addends addends
num_sums = []
# We iterate the number of addends from 0 to 49, not including 49:
for num_addends in range(49):
# Append an array into num_sums for num_sums[num_addends]:
num_sums.append([])
for sum in range(1025):
# If there are no addends,
# then there is 1 way to add to 0
# and 0 ways to add to anything else:
if num_addends == 0:
num_sums[num_addends].append(1 if sum == 0 else 0)
# Otherwise:
else:
# At first, we have found 0 ways:
num_sums[num_addends].append(0)
# Loop through all the addends:
for addend in addends:
# If addend is bigger than sum,
# exit because that means there is no way
# to include addend in our sum:
if addend > sum: break
# Increment num_sums[num_addends][sum]
# by the number of ways that sum-addend can be found
# with num_addends-1 addends:
num_sums[num_addends][sum] += \
num_sums[num_addends-1][sum-addend]
# Output the number of ways to sum to 3 GB with 2 addends.
# It is easy to show that this should be 2,
# so we know we did something wrong if our program does not output 2:
print(num_sums[2][3])
# Output the number of ways to sum to 1024 GB with 48 addends.
print(num_sums[48][1024])
Also, please note that my program accounts for the order in which the slots are filled , so the situation where $1$ GB is put in the first RAM slot and $2$ GB is put in the second RAM slot and the situation where $2$ GB is put in the first RAM slot and $1$ GB is put in the second RAM slot are treated as the different situation and are counted as $2$ ways to sum to $3$ with $2$ addends. Also, I leave no RAM slot empty, so each RAM slot has at least $1$ GB of RAM.
My program outputs $54725807780812926153007692867349522953$ as an answer, meaning that there are $54725807780812926153007692867349522953$ ways to sum to $1024$ GB using $48$ RAM slots with either $1$, $2$, $4$, $8$, $16$, $32$, and $64$ GB of RAM while account for the order in which the RAM slots are filled and leaving no RAM slot empty.
The answer depends on whether you require each of the six flavours to be present (that is, at least 1%).
Assuming that you do indeed intend each flavour to be present: Consider 100 little cups waiting to take their flavour portion, representing 1% each. Now insert marks between the cups to show where to change to the next flavour - you need 5 such marks, to insert into 99 different positions.
Thus the answer is $\binom {99}{5} = \frac{99!}{94!\cdot5!}$.
For the other case where some flavours can be absent, this means you can have multiple marks in one space. Now the result looks like arrange 100 cups and 5 marks, which means we have 105 places and we choose 5 of those places to be flavour-change marks.
For this case the answer is $\binom {105}{5} = \frac{105!}{100!\cdot5!}$.
Best Answer
From a comment by Mike Earnest: