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Q2RE

Expert-verifiedFound in: Page 439

Book edition
7th

Author(s)
Kenneth H. Rosen

Pages
808 pages

ISBN
9780073383095

**Explain how to find the number of bit strings of length not exceeding 10 that have at least one 0 bit.**

The number of string of length $\le 10$ having at least 0 bit is 2036 .

**Permutations: A permutation of a set is a loosely defined arrangement of its members into a sequence or linear order, or, if the set is already ordered, a rearrangement of its elements, in mathematics. The act of changing the linear order of an ordered set is also referred to as "permutation."**

** **

**Lexicographic order: The lexicographic or lexicographical order (also known as lexical order or dictionary order) in mathematics is a generalization of the alphabetical order of dictionaries to sequences of ordered symbols or, more broadly, elements of a totally ordered set. **

Considering the given information:

Length of string $\le 10$ having at least one 0 bit.

Using the following concept:

Number of bit string of length $n={2}^{n}$

The number of bit strings of length $\le 10$ can be calculated as:

$={2}^{1}+{2}^{2}+{2}^{3}+{2}^{4}+{2}^{5}+{2}^{6}+{2}^{7}+{2}^{8}+{2}^{9}+{2}^{10}\phantom{\rule{0ex}{0ex}}=2+4+8+16+32+64+128+256+512+1024\phantom{\rule{0ex}{0ex}}=2046$

Number of strings without zeroes of length 1=1

Number of strings without zeroes of length 2=1

Number of strings without zeroes of length 3=1

Number of strings without zeroes of length 10=1

As a result, the number of bit strings of length $\le 10$ with at least one 0.

=2046-10

20336

Therefore, the required number of string of length $\le 10$ having at least 0 bit is 2036 .

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