Least number of elements that combined form 7.8 billion different combinations

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Meslor

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Hi everyone,

I'm not sure if this is the correct section, in case it's not, i apologise in advance.
As the title says, what's the least number of elements that combined form 7.800.000.000 different combinations, and what is the formula to calculate it?

Thank you in advance.
 
Meslor said:
Hi everyone,

I'm not sure if this is the correct section, in case it's not, i apologise in advance.
As the title says, what's the least number of elements that combined form 7.800.000.000 different combinations, and what is the formula to calculate it?

Thank you in advance.
Click to expand...
Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this problem
 
Thank you for the reply. First, this is a problem that came up to my mind and it's not an homework.
I'm stuck at the very beginning of the process because i cannot figure out how to approach this problem.
Firstly i thougth that maybe the square root of 7.800.000.000 could be the solution, but i don't think it's correct.
Then, i thougth that a number exponentiated by itself could be the solution, but i don't know how to reverse an exponentiation.
As you can see, i'm not a mathematician, therefore i'm asking for your help, i only hope this question is compliant with the rules.
 
Meslor said:
Hi everyone,

I'm not sure if this is the correct section, in case it's not, i apologise in advance.
As the title says, what's the least number of elements that combined form 7.800.000.000 different combinations, and what is the formula to calculate it?

Thank you in advance.
Click to expand...
Please clarify what you mean. When you aren't sure of the right mathematical terms, you need to use more words to be clear.

In what way are you "combining" the elements? An example will help; show us some (few) elements and how many combinations you get. "Combinations" is a technical term with a precise meaning, but it sounds like you don't know whether it is the right term for what you want.

It may also help to tell us why you are asking; the application will determine the math.
 
There are multiple common ways to answer such questions.

Let’s say we have four distinct elements a, b, c, and d, and we want know what happens if two are selected.

aa, ab, ac, ad, ba, bb, bc, bd, ca, cb, cc, cd, da, db, dc, dd

16 ways to ORDER two out of four items if an item can be selected more than once (with replacement).

[MATH]16 = 4^2[/MATH]
ab, ac, ad, ba, bc, bd, ca, cb, cd, da, db, dc

12 ways to ORDER two out of four items if an item cannot be selected more than once (without replacement).

[MATH]12 = \dfrac{4!}{2!}[/MATH]
ab, ac, ad, bc, bd, cd

6 ways to SELECT, without replacement and without regard to order, two out of four items

[MATH]6 = \dfrac{4!}{2! * (4 - 2)!}[/MATH]
Three different questions. Three different formulas. What is YOUR question?
 
\( 7,800,000,000 = 2^9 \times 3 \times 5^8 \times 13 \), these are the prime factors. It seems strange that 7 and 11 are missing. But you could fabricate a question that results in this. How many different results could you obtain after performing ALL the following:-
  • take 8 numbers from a bag containing balls numbered 1-10 with replacement, and the order matters (10^8 ways)
  • toss a coin (2 ways)
  • roll a 13 sided die (13 ways)
  • catch an apple on a trident, which of the three spikes pierces the apple (3 ways)
(10)^8 * 2 * 13 * 3 = 7800000000

...there are many other possibilities. I can't think of a very simple event that would have this number of outcomes. A die with 7800000000 sides surely doesn't exist? However, who knows the lengths that dungeons and dragons game players will go to ?
 
Meslor said:
I'm not sure if this is the correct section, in case it's not, i apologise in advance.
As the title says, what's the least number of elements that combined form 7.800.000.000 different combinations, and what is the formula to calculate it?
Click to expand...
Another aspect that needs clarification is, do you want exactly 7.800.000.000 different combinations, or at least 7.800.000.000 different combinations? In the former case, factors matter; in the latter, they don't. We really, really need to know more about the goal, which might include, "Why that number??"

Just to suggest the simplest possibility, suppose that you are asking about choosing M items from N types (like 3 scoops of ice cream out of 27 flavors), allowing repetition but distinguishing order (chocolate, vanilla, chocolate is one). Then there are N^M ways to choose and order the items. Then if [MATH]N = 100[/MATH], M has to be at least [MATH]\log_{100}7 800 000 000 = 4.94[/MATH]; so you could choose 5 items of 100 flavors. If [MATH]N = 50[/MATH], M is at least [MATH]\log_{50}7 800 000 000 = 5.82[/MATH], so you could choose 6 items of 50 flavors.

Is that anything like what you are asking???
 
Thank you so much for your kind replies.
Okay, i'll try to explain better:
Let's say, hypothetically, that i want to create exactly 7.8 billions different and unique graphical symbols. Let's say that, because it's impossible to manually draw these symbols, i need to program an automatic generator that would combine a set of basic graphical elements (items) to make those 7.8 billions symbols. The program has to be the most efficient as possibile, so i need to know what's the least number of items i must have to create the number of combinations as said above. The items can be selected more than once with replacement, so i think JeffM answered my question with the First example.
I hope i've sufficiently explained my question (i'm not native english), so:

16 ways to ORDER two out of four items if an item can be selected more than once (with replacement).

if 16=4²

Therefore

7800000000 = x²

Therefore
√7800000000=x

Is this the answer?

Thank you so much
 
Meslor said:
16 ways to ORDER two out of four items if an item can be selected more than once (with replacement).

if 16=4²

Therefore

7800000000 = x²

Therefore
√7800000000=x

Is this the answer?
Click to expand...
This shows that there are 780 000 000 ways to fill in two blanks, _ _, with any of about 27 928 items.

Surely you don't want your symbols to each be made of only 2 items, do you?

Let's suppose that a symbol consists of m locations, each of which can hold any of n possible elements. Then the number of possible symbols will be n^m. This is identical to what I said about ice cream! So if a symbol consists of 10 locations, the number of elements you need to make more than 780 000 000 symbols is 780 000 000^1/10 = 7.74, since you want m^10 = 780 000 000; you could use 8 symbols and have extra symbols you won't use (a total of 1 073 741 824 symbols).

But to make it exactly 780 000 000, you need 780 000 000 to be a perfect power of some integer base, and you have already been told that
\( 7,800,000,000 = 2^9 \times 3 \times 5^8 \times 13 \), which is not such a power. So your goal is impossible.
 
what's the least number of elements that form 7.800.000.000 different combinations?

I would say 33.
There are 33 combinations that have just one element
(33 x 32) / 2 = 528 that have two elements
(33 x 32 x31) / (3 x2) = 5.456
and so on
Total number of combinations n^33 = 8,589,xxx,xxx

Devise 33 symbols (graphical elements) 1 - 33
Write your program to produce strings of length 33
000 ... 000
000 ... 001
000 ... 010
000 ... 011
... ... ...
111 ... 100
111 ... 101
111 ... 110
111 ... 111
"0" in positon n to mean no symbol in position n
"1" in postion n to mean the nth symbol in position n
This is more than you need so your program would stop when 7,8000,000,00 is reached.
Alternatively, start at 111 ... 111 and work down, stopping after 7,800,000,000.
 
1621935320005.png
The logo might have 19 positions, and use 12 elements: A to L
(If you don't like empty you can add a letter instead).
This gives 7.8 billion combinations.
I don't know if this is satisfactory!
 
lex said:
View attachment 27458
The logo might have 19 positions, and use 12 elements: A to L
(If you don't like empty you can add a letter instead).
This gives 7.8 billion combinations.
I don't know if this is satisfactory!
Click to expand...

..OP, you can obviously substitute a graphic element for any of the above boxes. For example empty could be represented by an upward squiggle, "^", and A could be represented by a downward squiggle, "v". So the boxes 1-9 could be shown as a wave-like line. Or, one of the boxes could represent whether or not an eyebrow is raised on a depiction of a face.

You can combine boxes too, if that suits your requirement. This means there will be fewer boxes, but the combined boxes will have more options. For example, you could combine box 1 & 2 (which each have two options) into a single box, let's call it "1&2", that has 2*2=4 options:- blank blank; A blank; blank A; or A A. These options may be drawn as "the sun", "the moon", "saturn", or "shooting star", for example.

There are lots of options to consider!
 
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