# How many $8$-digit numbers formed using the digits from $1$ to $9$ have exactly $5$ consecutive even digits and exactly two odd digits?

Suppose an $8$-digit number will be formed using the digits from $1$ to $9$. Find the cardinality when exactly $5$ consecutive digits are even and exactly $2$ digits are odd, e.g., $18242674$, $26682523$

I thought of dividing it into four subgroups, the five consecutive even integers, the two odd numbers, and the other even integer. ie, $(4!) (4)^5 \times (5)^2 \times (4)^1$. I plan to use this cardinality for the classical approach of a prob. Thanks.

• Well, there aren't very many places to put the block of $5$. Just count each possible pattern. – lulu Oct 2 '17 at 13:08
• Your examples have only one odd digit. – N. F. Taussig Oct 2 '17 at 14:14
• There are two odd digits in each example – J. Gielgud Oct 2 '17 at 16:38

Such numbers take one of the following four forms: $$\mathtt{EEEEEOXX} \quad \mathtt{OEEEEEOE} \quad \mathtt{EOEEEEEO} \quad \mathtt{XXOEEEEE}$$ where $\mathtt{E}$ denotes an even digit, $\mathtt{O}$ denotes an odd digit, and $\mathtt{XX}$ denotes a pair of digits in which one is even and one is odd.
• Perhaps $\mathtt{EEEEEOOE}, \mathtt{EEEEEOEO}, \mathtt{OEEEEEOE}, \mathtt{EOEEEEEO}, \mathtt{EOOEEEEE}, \mathtt{OEOEEEEE}$ is easier since they all have equal cardinality – Henry Oct 2 '17 at 13:28