Given that there are 32 eight-bit strings that begin 101 and 16 eight-bit strings that begin 1101, how many eight-bit strings begin either 101 or 1101?

Question
Given that there are 32 eight-bit strings that begin 101 and 16 eight-bit strings that begin 1101, how many eight-bit strings begin either 101 or 1101?

Answers (1)

2021-02-26
Let A be the set of all eight-bit strings that begin with 101 and B be the set of all eight-bit strings that begin with 1101. Then
n(A)=32, n(B)=16
(n(X) gives us the number of elements of X). Since A and B are disjoint, we get that \(\displaystyle{n}{\left({A}∪{B}\right)}={n}{\left({A}\right)}+{n}{\left({B}\right)}={32}+{16}={48}\),
which is our answer.
48
0

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