Let a good-block be a block that is $ME$ of $MO$. Consider all strings that have exactly $x$ good-blocks. If $x$ is odd, the string cannot have an even number of both $ME$ and $MO$ blocks, or an odd number of both $ME$ and $MO$ blocks. Thus it contributes a value of $0$ to both $a$ and $b$. If $x>0$ is even, these strings contribute a value of $\binom{x}{0} + \binom{x}{2} + \binom{x}{4} + \cdots$ to $a$, and a value of $\binom{x}{1} + \binom{x}{3} + \binom{x}{5} + \cdots$ to $b$. It is well known that these sums are equal, so all strings with $x$ being even contribute an equal amount to $a$ and $b$. Now we are left with the $x=0$ case. All such strings contribute to $a$, but not to $b$. As there is at least one such string, such as $EE \cdots E$, we can conclude that $a > b$.

We can even compute $a-b$. According to the above, it is the number of $n$-strings having no good-block ($ME$ or $MO$). If the first $M$ occurs on some position, all further positions must be $M$. Therefore $a-b = 2^n+2^{n-1} + \cdots + 2 + 1 = 2^{n+1} - 1$.

another solution: let $A$ be the set of words that have even number of $ME,MO$ and $B$ be the set of the words that have odd number of $ME,MO$.we establish an one to one correspondence between elements of $B$ with elements of $A$ in such a way that for every word in $B$ consider the first $ME$ or $MO$ (from left to right) .if it was $ME$ change it to $MO$ and if it was $MO$ change it to $ME$.now we get a word in $A$.it's obvious that the correspondence is one to one so $\left |A \right | \geq \left |B \right |$ and because we don't correspond any word in $B$ to the word $MM...M$ that occur in $A$ so $A>B$