How many sets $A \subset [1,2,...,100]$ of size $20$ satisfy $|a_1 - a_2|\geq 2$ for distinct $a_1, a_2 \in A$ I've got the next interval: $[1,2,...,100]$. How many sets of $20$ the subtraction between each two numbers in the set is at least $2$?
I've got a starting idea simplifying the question [Might be incorrect]: 
We'll look at the question this way, I've got a set of $20$ cells with $19$ partition between them and $n$ amount of balls. [We'll look at $n$ soon enough]. The cells will represent the numbers $1,2,\ldots,20$ and the partitions will represent the 'space' between the numbers. [This means, I'll add the partition to the right cell]. For the first number of the set we'll add a partition grouping up with $1$ [Though, it won't be a part of the set].
So, given that idea, and since the minimum amount of subtraction between two cells is $2$, I'll insert $1$ ball into each partition [total number of $19$ balls].
Summary: Say, we have $1,3,5,7, \ldots, 39$, means it looks like: $$1||1|o|2|o|3|o|5|...|o|20$$
Looking at $n$, it's maxed by $80$ [Since if I'll add up $80$ to $20$, it'll be $100$], and since I already used $19$, I'm left with $61$ balls. So, the number of possible options is just like looking at an equation like this:
$$x_1+x_2+x_3+\ldots+x_{20}=k,$$ where $x_1,x_2, \ldots$ represents the partitions looking from left to right, and $k$ represents the number of balls left.
So I get the sum of $\binom{20+k-1}{20-1}$ from $k=0$ to $k=61$.
Am I right or am I missing something?
 A: The basic idea is OK, but adding two balls between numbers forces them to have distance $\gt2$, not $\ge2$. So you only want one fixed ball between any two numbers. That leaves $100-20-19=61$ balls, which can be distributed arbitrarily over $21$ different slots.
I'm not sure I understand the second part of your approach; as far as I understand it it looks wrong. Using the stars and bars approach, the number of ways to distribute $61$ balls over $21$ slots is found to be $\displaystyle\binom{61+21-1}{61}=\binom{81}{61}$.
A: I would just like to note that using a start-and-bars like approach is really not necessary. The requirement that no two adjacent numbers can be chosen means that every chosen number must be followed by a non-chosen number, except for the very last chosen number where this is not required. So you can treat each of the first $19$ chosen numbers together with the following un-chosen number as a single unbreakable chosen unit (the final chosen number forms a chosen unit in its own). This pairing up effectively reduces the number of elements by $19$, so the problem is equivalent to choosing $20$ elements in a list of $100-19=81$. You could similarly handle any minimum distance requirement: if $k$ elements out of $n$ have to be chosen with minimal distance$~d$ between them, there will be $\binom{n-(k-1)(d-1)}k$ possibilities.
