This is a question about a personal problem that I have been trying to solve, which concerns combinations and summations,
The problem, in summary, deals with $10^8$ objects, and finding the total amount of ways in which they could be combined, where order does not matter, and repetition is not allowed. This should come out to
$$ \sum_{r = 1}^{10^8}\frac{10^8!}{r!(10^8 - r)!}, $$
I do believe I have found $(10^8)!$ as ~$2.5\cdot(10^{75670555.81})$, but I cannot figure out how to solve the rest of the problem, from a mixture of not being able to find calculators that can deal with such quantities, and due to not having the millennia needed to do it by hand.
Is there any other easier way to do this? To at least get an approximation to it?
I am aware larger numbers have been calculated, so I do not see finding an answer as being too far out of the question. Also, do not fret with complicated answers, I find myself relatively well-versed in mathematics and can handle advanced methods.
Observe that your sum is equivalent to $$ \sum_{r = 1}^{10^8}\binom{10^8}{r}. $$
The binomial theorem tells us that for any nonnegative integer $n,$ we have $$ (x + y)^n = \sum_{i = 0}^n\binom{n}{i}x^i y^{n - i}. $$
Thus, it follows that $$ (1 + 1)^n = 2^n = \sum_{i = 0}^n\binom{n}{i}. $$ In particular, your sum becomes $$ \sum_{r = 1}^{10^8}\binom{10^8}{r} = \left(\sum_{r = 0}^{10^8}\binom{10^8}{r}\right) - 1 = 2^{10^8} - 1. $$
Now, all you need to do is find a calculator which can approximate this very large number. Wolfram Alpha approximates this as $$ 3.684665936980458763209092390984221915069965812267549708 \times 10^{30102999} $$ or $$ 10^{10^{7.478609772345674}}. $$
As a note, it seems you can count what you want by realizing that you want to count the number of subsets of a set with $10^8$ elements which have at least one element. The number of subsets of a set with $10^8$ elements is $2^{10^8}$ (for each element, it can either be included or left out), and the only subset with no elements is $\emptyset$.