Choosing books on a bookshelf where no two are consecutive I'm going through Cohen's Basic Techniques of Combinatorial Theory and I am stuck on this problem:
There are 24 volumes of an Encyclopedia on a bookshelf. In how many ways can 5 of these books be selected if no 2 consecutive volumes are to be chosen?
Any help would be much appreciated! 
Thanks  
 A: I love to think about this problem this way:  
We start choosing by the lowest number, and will not choose any number before that number and I proceed by choosing only in increasing order.  
So, when we choose the lowest number, we eliminate the number next to the chosen number. We continue this process and on the final selection of number, We have chosen all $5$ numbers so we do not have to eliminate any number anymore.  
If one thinks carefully, then one will observe that we have eliminated $4$ numbers and we can choose any $5$ numbers from the remaining $20$ numbers as $20\choose 5$
A: You have to count the number of binary strings with $24$ bits and $5$ non-consecutive ones. This can be done by taking $19$ consecutive zeroes and by choosing which "gaps" between the zeroes (including the gap preceding the first zero, and the gap following the last one) have to be filled by ones. Since there are $20$ gaps and you have to fill $5$ gaps, the answer is just $\binom{20}{5}$.
A: Hint: try to find the equivalence to the following question: "There are 20 volumes of an Encyclopedia on a bookshelf. In how many ways can 5 of these volumes be selected? (with no special restriction)"
A: As the numbers are non-consecutive, so if we select 1 number, we can skip any amount of numbers (atleast 1 should be skipped) after that number, and before choosing the first number we may or may not skip the numbers. Continuing this way until the 5th number, after that we may be having any amount of non-choosen numbers, which accounts to total 19 non-chosen numbers, 
So, this now becomes same as
$\text{coefficient of $x^{19}$ in the expansion of}$
$$(1+x^2+x^3+\cdots)^2\cdot (x +x^2+x^3+\cdots)^4$$
$$=x^4(1-x)^{-6}$$
$\text{= coefficient of $x^{15}$ in $(1-x)^{-6}$}$
$$=\binom{6+15-1}{15} = \binom{20}{15} = \binom{20}{5}$$
