The question : How many bit strings of length 10 either begin with three $0$s or end with two $0$'s?
My solution : $0$ $0$ $0$ X X X X X $0$ $0$ = $2^5 = 256$
editing** I noticed the word"or" so I changed the solution to
$2^7$ (three $0$'s) +$2^8$(two $0$'s) – $2^5$(both) =416
is this the correct way to do it?
Best Answer
These are the binary words $000x$ ($2^7$ many $x$) and $y00$ ($2^8$ many $y$) minus $000z00$ ($2^5$ many $z$). Looks good.
But I calculate $352$.