A question about a simple combinatorial problem I am reading "Introduction to Combinatorial Mathematics" by C. L. Liu.
And I cannot understand the meaning of "because of symmetry" in the following sentences:
p.7 Example 1-9
"A binary sequence is a sequence of $0$'s and $1$'s. What is the number of $n$-digit binary sequences that contain an even number of $0$'s(zero is considered as an even number) ? The problem is immediately solved if we observe that because of symmetry half of the $2^n$ $n$-digit binary sequences contain an even number of $0$'s, and the other half of the sequences contain an odd number of $0$'s."
Liu says that "because of symmetry".
In fact, if $n$ is an odd integer, the problem is immediately solved by symmetry as follows(but if $n$ is an even integer, the following argment doesn't work):
If a sequence contains an even number of $0$'s, then the sequence contains an odd number of $1$'s.
If a sequence contains an odd number of $0$'s, then the sequence contains an even number of $1$'s.
So, "the number of sequeces which contain an even number of $0$'s" is equal to "the number of sequeces which contain an odd number of $1$'s",
and, "the number of sequeces which contain an odd number of $0$'s" is equal to "the number of sequeces which contain an even number of $1$'s".
Obviouly, "the number of sequeces which contain an even number of $0$'s" $+$ "the number of sequeces which contain an odd number of $0$'s" is equal to $2^n$.
And, by symmetry, "the number of sequeces which contain an even number of $0$'s" is equal to "the number of sequeces which contain an even number of $1$'s".
Finally,
$2^n$
$=$
"the number of sequeces which contain an even number of $0$'s" $+$ "the number of sequeces which contain an odd number of $0$'s"
$ = $
"the number of sequeces which contain an even number of $0$'s" $+$ "the number of sequeces which contain an even number of $1$'s"
$ = $
"the number of sequeces which contain an even number of $0$'s" $\times 2$.
So, 
"the number of sequeces which contain an even number of $0$'s" is equal to $2^{n-1}$.

Let the number of binary sequences of $n$ digits such that the number of $0$s is even be $E_n$, and similarly define $O_n$ for an odd number of $0$s, with $E_n + O_n = 2^n$. Looking at the sequences of length $n-1$, there are $E_{n-1}$ that have an even number of $0$s, and $O_{n-1}$ that have an odd number.
I then claim that there are four groups of sequences of length $n$:


*

*Sequences where the first digit is $0$, and the remaining sequence of length $n-1$ has an even number of $0$s, such that the whole sequence has an odd number of $0$s (there are $E_{n-1}$ of these).

*Sequences where the first digit is $1$, and the remaining sequence has an even number of $0$s, such that the whole sequence has an even number of $0$s (there are $O_{n-1}$ of these).

*Sequences where the first digit is $0$, and the remaining sequence has an odd number of $0$s, so the whole sequence has an even number (there are $E_{n-1}$).

*Sequences where the first digit is $1$, and the remaining sequence has an odd number of $0$s, so the whole sequence has an odd number (there are $O_{n-1}$).


So the sequences with an even number of $0$s come from groups 2 and 3, while the ones with an odd number come from groups 1 and 4. But the total size of groups 2 and 3 is $E_{n-1} + O_{n-1}$, as is the size of groups 1 and 4, so the two groups are of equal size, and hence $E_n = O_n$.
