How many bit strings of length 10 either begin with three 0s or end with two 0s?
I solved this question using cases but I do not seem to be getting the answer of $352$.
My attempt:
Consider two cases:
- Case 1: The string begins with three $0$s and does not end with two $0$s. There is only $1$ way to choose the first three bits, $2^5$ ways for the middle bits, and $3$ ways for the last two bits ($4$ ways to construct a string of two bits, minus $1$ way to make three $0$s). There are $2^5 \cdot 3$ ways to construct strings of this type.
- Case 2: The string does not begin with three $0$s but ends with two $0$s. There are $2^3 – 1 = 7$ ways to choose the first three bits without three $0$s, $2^5$ ways for the middle bits, and $1$ way for the last two bits. There are $7 \cdot 2^5$ ways to construct strings of this type.
By the rule of sum, there are $2^5 \cdot 3 + 2^5 \cdot 7 = 320$ ways to construct bit strings of length 10 either begin with three $0$s or end with two $0$s.
Best Answer
You are missing the strings that both begin with three zeroes and end with two zeroes.
And since there are five bits left that can be anything, you have $2^5=32$ of those, exactly the difference