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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
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Verified Answer
The correct answer is:
$2^{n-1}$
The required number of ways $=$ The even number of 0 's ie, $\{0,2,4,6, \ldots\}$
$\begin{aligned}
& =\frac{n !}{n !}+\frac{n !}{2 !(n-2) !}+\frac{n !}{4 !(n-4) !} \\
& ={ }^n C_0+{ }^n C_2+{ }^n C_4+\ldots=2^{n-1}
\end{aligned}$
$\begin{aligned}
& =\frac{n !}{n !}+\frac{n !}{2 !(n-2) !}+\frac{n !}{4 !(n-4) !} \\
& ={ }^n C_0+{ }^n C_2+{ }^n C_4+\ldots=2^{n-1}
\end{aligned}$
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