I've been asked to find the number possibilities with a 12 bit string, where there are exactly four 1s, where none are adjacent to each other.
I'm pretty sure that if the adjacent property wasn't necessary, then it would be 12!/(4!*8!), but I'm puzzled on how to solve this.
Best Answer
Write down the eight $0$'s, with some space between them, like this: $$0\qquad 0\qquad0\qquad0\qquad0\qquad0\qquad0\qquad0\qquad$$ There are $9$ "gaps" determined by these $0$'s ($7$ real gaps, plus just before the first $0$, and just after the last one).
We need to choose $4$ of these gaps to put a $1$ into. By definition, there are $\binom{9}{4}$ ways to do this.
Remark: The number $\binom{9}{4}$ denotes a binomial coefficient. You may be accustomed to calling it ${}^9C_4$, or $C^9_4$, or $C(9,4)$, or by some related name. It is equal to $\frac{9!}{4!5!}$, which simplifies to $\frac{(9)(8)(7)(6)}{4!}$, and then to $126$.