How many ways can $r$ nonconsecutive integers be chosen from the first $n$ integers? During self-study, I ran across the question of how many ways six numbers can be chosen from the numbers 1 - 49 without replacement, stipulating that no two of the numbers be consecutive.
I can obtain a simple lower bound by saying that, in the worst-case scenario, when you choose a particular number, there are now three numbers that cannot be chosen next.  For example, if I first pick 17, then I can't choose 16, 17, or 18 for the remaining choices.  This gives me the lower bound
$$\frac{49*46*43*40*37*34}{6!} = 6,773,770 \frac{8}{9}$$
This is about 48% of ${}_{49}C_6 =  13,983,816$.  The real answer must be bigger (and an integer).  I haven't found a way to calculate it, though.
The original problem asked to show that the probability of having non-consecutive integers when you choose six is greater than 50%, so if the problem is complicated to count exactly, better approximations that show the answer is above 50% would also be appreciated. 
Of course, I can use a computer to do the counting, but I'm interested in learning what methods I'm missing.
 A: In analogy to the Pascal triangle, define D(n,r) as the number of ways to select r items from n with no two consecutive.  You can either use 1 or not.  If you do use it, you can't use 2 but only need to pick r-1.  If you don't use it you have n-1 items to pick r from.  So D(n,r)=D(n-2,r-1)+D(n-1,r).  The base case is D(2r-1,r)=1For values like this you can use a spreadsheet.
A: Another good way of solving problems like this is to set up a correspondence with a problem you already know how to solve.  In this case, imagine that you've got a sorted set of six non-consecutive numbers $a_1, a_2, \dots a_6$ between 1 and 49.  What does it look like?  Well, the first number $a_1$ is essentially arbitrary; it can't be greater than a certain value (i.e. 39) or else there isn't 'room' for five more non-consecutive numbers above it, but we can ignore that for now.  The second number $a_2$ has to be greater than $a_1+1$ — that's what it means to be nonconsecutive, after all — so we know that $a_2-1$ (call this $b_2$) is greater than $a_1$.  Similarly, $a_3$ has to be greater than $a_2+1$, so $a_3-1$ is greater than $a_2$, and $a_3-2$ is greater than $a_2-1 = b_2$; we can call this $b_3$.  And so on, and so forth, until we define $b_6 = a_6-5$.  But this correspondence works both ways — given the $b_n$ it's easy to get the $a_n$ — and so we have a one-to-one correspondence between our non-consecutive sets of numbers and an ordinary combination (but with a different upper bound - can you see what it should be?).  It takes a while to build this sort of instinct, but learning how to spot these correspondences is the most basic tool for proving most combinatorial identities.
