Search any question & find its solution
Question:
Answered & Verified by Expert
A binary sequence is an array of 0 's and l's. The number of $n$-digit binary sequences which contain even number of 0 's is
Options:
Solution:
2223 Upvotes
Verified Answer
The correct answer is:
$2^{n-1}$
The required number of ways $=$ The even number of $0^{\prime}$ s i.e., $\{0,2,4,6, \ldots\}$
$$
\begin{array}{l}
=\frac{n !}{n !}+\frac{n !}{2 !(n-2) !}+\frac{n !}{4 !(n-4) !} \\
={ }^{n} C_{0}+{ }^{n} C_{2}+{ }^{n} C_{4}+\ldots=2^{n-1}
\end{array}
$$
$$
\begin{array}{l}
=\frac{n !}{n !}+\frac{n !}{2 !(n-2) !}+\frac{n !}{4 !(n-4) !} \\
={ }^{n} C_{0}+{ }^{n} C_{2}+{ }^{n} C_{4}+\ldots=2^{n-1}
\end{array}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.