In how many 7-digit numbers $5$ threes came one after another? In how many 7-digit numbers $5$ threes came one after another?
I put one $3$ for all $5$ threes after thiswe know that this is the case of three digit numbers that a $3$ always exist now For easy solving I calculate the case that never a $3$ exist.Now by calculating I get this answer:
$900-8*9*9=900-648=252$
But the book give the answer $651$ where did I do wrong?
 A: I'm thinking the book is answering a different question.  (The $648$ in your answer is very close to $651$, the answer the book gives, so maybe something is getting lost in translation.)
However, the $252$ you have doesn't quite account for everything.  You're using a single $3$ to represent the five $3$s in a row, which is fine, but you're missing some cases.
You're taking the $900$ three-digit numbers from $100$ to $999$, and calculating how many of those don't contain any $3$s.  You did this correctly.
The question this answers, though, is:

"How many seven-digit numbers (with leading zeroes not allowed)
  contain exactly five $3$s, all in a row?"

This is a different question than what appears to have been asked:

"How many seven-digit numbers (with leading zeroes not allowed)
  contain five $3$s in a row?"

For the second question, $9333333$ works.  There's more than five $3$s in a row, but that's OK.  Also, $3733333$ works.  There's another $3$ at the beginning, but there's still five $3$s in a row.  Even $3333333$ works!
What Ross's answer shows you is how to include these extra cases.  
A: You have three possible patterns:  $ab33333, a33333b,$ or $33333ab$.  Each of the first two gives $90$ possibilities while the last gives $100$, for a provisional total of $280$.  Now you need to subtract the cases with six $3$s in a row because you have counted them twice, of which there are $19$.  The case of seven threes has been counted three times in the first and subtracted twice in the second, so we have one like we should.  $280-19=261$ for a final answer.  The book is either wrong or answering a different question.
A: The string of five consecutive $3$'s can start with the first, second, or third digit.  In the first case, the final two digits can be anything, so it accounts for $10\cdot10=100$ numbers.  In the second case, the first digit can be anything but a $0$ or a $3$ while the final digit can be anything, so it accounts for $8\cdot10=80$ numbers.  In the third case, the first digit can be anything but a $0$ and the second digit can be anything but a $3$, so it account for $9\cdot9=81$ numbers.  We get a total of
$$100+80+81=261$$
seven-digit numbers with five consecutive $3$'s.
