From Richard Stanley's Enumerative Combinatorics,
2(j) How many sequences $(a_1, a_2, \ldots, a_{12})$ are there consisting of four 0's and eight 1's, if no two consecutive terms are both 0's?
I can brute force this, but Stanley provides the (mysterious) reasoning ${8 + 1 \choose 4} = 126$.
Any elegant argument is appreciated.
From the answer here: Consecutive birthdays probability
We need to select $\displaystyle d$ numbers from $\displaystyle 1,2, \dots, k$ such that no two are consecutive.
Now if $\displaystyle b_1 \lt b_2 \lt \dots \lt b_d$ were such numbers, then notice that
$\displaystyle 1 \le b_1 \lt b_2 - 1 \lt b_3 - 2 \dots \lt b_d - (d-1) \le k-(d-1)$ gives us a way to select numbers from $\displaystyle 1, 2, \dots, k-(d-1)$ without having to bother about the consecutive issue.
This can be done in $\displaystyle {k-d+1 \choose d}$ ways.
In your case, $k=12, \ d=4$.
Let me attempt a "proof without words":
_ 1 _ 1 _ 1 _ 1 _ 1 _ 1 _ 1 _ 1 _
(There are nine slots here. Pick four of them to put zeroes in.)
