Messaging probabilities I am part of a large family - we have twenty-four people who send texts back and forth, in various configurations. What would be the total number of possible message threads? All the different one-on-ones - the three persons groups, four, etc., up to all twenty-four of us on one thread.
 A: $\require{cancel}$
Alternatively, suppose that we have a thread that only $2$ members participate. Because order doesn't matter, we have $\dbinom{24}{2}$ such threads. The number of threads that any $3$ members participate is $\dbinom{24}{3}$. Following the same procedure, we have that the total amount of threads is:
$$\sum_{n=2}^{24} \dbinom{24}{n}=16,777,191,$$
which agrees to ToolPurger's answer.

We can work the same way, as we did before. Suppose that we fix a person (yourself), which we want to participate in every thread. Let's begin with the thread with $2$ participants (the one is yourself). Thus, we have to pick the other person from the rest $23$ people. We can do that with $\dbinom{23}{1} = 23$ ways. The number of threads with $2$ people (again, the one is yourself) is $\dbinom{23}{1}$. 
Moving to the thread with $3$ participants (the one is yourself) means that we have to pick the $2$ participants from the rest $23$ people. We can do so with $\dbinom{23}{2}$ different ways. That means the number of threads consisting of $3$  people (again, the one is yourself) is $\dbinom{23}{2}$.
Following the same procedure, we have eventually that that number of threads you are included is:
$$\sum_{n=1}^{23}\dbinom{23}{n}=8,388,607.$$
 You can do the math .
A: If we encode:
$(0110\cdots)$ as 1st member (alphabetical ordering say) not included, 2nd and 3rd included, 4th not, so forth...
Then there are $2^{24}-25$ ways.
$2^{24}$ is the number of such binary strings, and 24 of those are messages to oneself, which we discount, as well as empty threads (messages to no one)

It is worth noting that the sums of  nCr and 2n are related, in fact:
  $$\sum^n_{i=0}\ ^nC_i=2^n$$
  So what this answer has is:
  $\sum^{24}_{n=2}\ ^{24}C_n=\sum^{24}_{n=0}\ ^{24}C_n -\ ^{24}C_0 - \ ^{24}C_1=2^{24}−1−24=2^{24}−25$ - as I claimed.

