How many bit strings of length seven either begin with two 0s or end with three 1s?

How Many Bit Strings Of Length Seven Either Begin With Two 0S Or End With Three 1S 1 The purpose of this question is to find the number of bit-strings of length $7$ beginning with two $0$s and ending with three $1$s. The sequence of binary digits is usually called a bit-string. The number of bits signifies the value length in the sequence. A bit-string having no length is regarded as a null string. Bit-strings are useful for representing sets and manipulating binary data. The bit-string elements are labeled left to right from $0$ to one minus the total number of bits in the string. When converting a bit string to an integer, the $0^{th}$ bit corresponds to the $0^{th}$ exponent of two, the first bit corresponds to the first exponent, and so forth. In discrete mathematics, the subsets are represented by the bit-strings in which $1$ indicates that a subset contains an element of a respective set and $0$ indicates that the subset does not contain that element. The representation of a set by a bit-string makes it simple to take complements, intersections, unions, and set differences.

Expert Answer

Let the set of bit-strings having the length $7$ and starting with two zeros be represented by $A$, then: $|A|=1*1*2*2*2*2*2=2^5=32$ Let the set of bit-strings having the length $7$ and starting with three ones be represented by $B$, then: $|B|=2*2*2*2*1*1*1=2^4=16$ Now, the set of the bit-strings of length $7$ starting with two $0$s and ending with three $1$s is given by: $|A\cap B|=1*1*2*2*1*1*1=2^2=4$ Finally, the number of bit-strings of length $7$ either starting with two $0$s and ending with three $1$s is: $|A\cup B|=|A|+|B|-|A\cap B|$ $|A\cup B|=32+16-4=44$

Example

How many numbers between $1$ to $50$ are divisible by either $2, 3$ or $5$? Assume $1$ and $50$ to be inclusive.

Solution

This example gives a clear idea of how the Sum Principle (Inclusion Exclusion) works. Let $A_1$ be the set of numbers between $1$ to $50$ which are divisible by $2$ then: $|A_1|=\dfrac{50}{2}=25$ Let $A_2$ be the set of numbers between $1$ to $50$ which are divisible by $3$ then: $|A_2|=\dfrac{50}{3}=16$ Let $A_3$ be the set of numbers between $1$ to $50$ which are divisible by $5$ then: $|A_3|=\dfrac{50}{5}=10$ Now, $A_1\cap A_2$ will be a set where each element between $1$ to $50$ is divisible by $6$, and so: $|A_1\cap A_2|=8$ $A_1\cap A_3$ will be a set where each element between $1$ to $50$ is divisible by $10$, and so: $|A_1\cap A_3|=5$ $A_2\cap A_3$ will be a set where each element between $1$ to $50$ is divisible by $15$, and so: $|A_2\cap A_3|=3$ Also, $A_1\cap A_2\cap A_3$ will be a set where each element between $1$ to $50$ is divisible by $30$, and so: $|A_1\cap A_2\cap A_3|=2$ Finally, using the sum principle to get the union as: $|A_1\cup A_2\cup A_3|=|A_1|+|A_2|+|A_3|-|A_1\cap A_2|-|A_1\cap A_3|-|A_2\cap A_3|+|A_1\cap A_2\cap A_3|$ $|A_1\cup A_2\cup A_3|=25+16+10-8-5-3+2$ $|A_1\cup A_2\cup A_3|=37$

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