How would I calculate the total number of combinations [closed]

Lets say I have 4 lines or rows lets call them Row 1 .. Row 4

Now the total number of ways to delete the rows are:

Row 1 (leaving Row2, Row3, Row4)
Row 2 (leaving Row1, Row3, Row4)
Row 3
Row 4
Row 1 and Row 2
Row 1 and Row 3
Row 1 and Row 4
Row 1 and Row 2 and Row 3
Row 1 and Row 2 and Row 4
Row 1 and Row 3 and Row 4
Row 2 and Row 3
Row 2 and Row 4
Row 2 and Row 3 and Row 4
Row 3 and Row 4

Now.. I need to know a mathematical formula that will allow me to calculate the total number of ways to delete the rows. In the provided example there are 14 different ways to delete. So how would I calculate if I for example have 924 rows? Please forgive me if I tagged this question incorrectly as I'm uncertain under which branch of mathematics this would fall under

closed as unclear what you're asking by JonMark Perry, C. Falcon, choco_addicted, user223391, ShaileshJun 25 '16 at 0:01

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• In your example, you are deleting some but not all the rows. If that's what you mean, then, given $n$ rows, there are $2^n-2$ ways to delete rows. – Barry Cipra Jun 24 '16 at 17:21
• Combinatorics and Elementary-Set-Theory were indeed the correct tags. Permutations, less so as this does not actually have anything to do with permutations. An equivalent way of describing the scenario, is that you have a set: $\{1,2,3,4\}$ and you are asking how many (proper and nonempty) subsets exist. The power set is the set of all subsets, and we know the power set of the set $S$ is of cardinality $2^{|S|}$. As mentioned, ignoring the case that we "delete everything" and "delete nothing" leaves us with $2^n-2$ (except in the case $n=0$) – JMoravitz Jun 24 '16 at 17:32
• Is there a condition that you must delete at least one row and not delete all the rows? If so, the number of ways to delete rows from 924 given rows is equal to the number of nontrivial (ie nonempty and proper) subsets of a 924 element set, which is equal to $2^{924}-2$. – Ashwin Ganesan Jun 25 '16 at 12:59
• Its unclear what I'm asking??? Have you Jon Mark Perry, C. Falcon, choco_addicted, Zachary Selk, Shailesh READ the question? Because others have and they ANSWERED it. Thank you to those that posted answers, those that put it on hold.. LEARN TO READ PROPERLY – Eminem Jun 25 '16 at 13:28

You left out two other ways to delete the rows: deleting none and deleting all of them, leaving a total of $16$ ways of deleting rows. Now you may notice $16=2^4$, and conjecture that they are $2^n$ ways to delete rows from a set of $n$. The general problem you are asking is how many subsets of a finite set of $n$ elements exist. (since the operation of deleting rows amounts to selecting some of them, i.e. a subset of the set of rows). In other words, you are asking how many elements are in the set of subsets of a set of size $n$. This is well known to be $2^n$.
To see this, notice there are $\binom{n}{k}$ ways of removing $k$ elements from a set of size $n$, so the total number of ways is $\sum_{k=0}^n \binom{n}{k}=(1+1)^n=2^n$. (applying $\sum_{k=0}^n \binom{n}{k}x^k=(1+x)^n$ with $x=1$) If you insist on leaving out trivial cases (removing no elements, yielding the original set, and removing all, yielding the empty set), then you get $2^n-2$.