Question

A binary sequence is an array of 0's and 1's. The number of \(n\)-digit binary sequences which contain even number of 0's is

• A. \(2^{n-1}\)
• B. \(2^n-1\)
• C. \(2^{n-1}-1\)
• D. \(2^n\)

Hint:

For binary strings, even and odd parity counts are equal, so each is \(2^{n-1}\).

Answer 1

The Correct Option is A

Step 1: Total binary sequences of length \(n\).
Each position has 2 choices (0 or 1), so:
\[ 2^n \]
Step 2: Use symmetry (even-odd parity).
In all binary sequences, number of sequences with even number of 0's equals number of sequences with odd number of 0's.
Because flipping the first bit maps even-zero sequences to odd-zero sequences one-to-one.
Step 3: Hence count of even-zero sequences.
\[ \frac{2^n}{2} = 2^{n-1} \]
Final Answer:
\[ \boxed{2^{n-1}} \]