Solution: 2016 Fall Midterm - 6

Author: Michiel Smid

Question

Let $n \geq 5$ and consider strings of length $n$ consisting of the characters $a$, $b$, $c$, and $d$. How many such strings are there that start with $ad$ or end with $dcb$?

(a)

$4^{n - 2} + 4^{n - 3}$

(b)

$4^n - 4^{n - 2} - 4^{n - 3}$

(c)

$4^n - 4^{n - 5}$

(d)

$4^{n - 2} + 4^{n - 3} - 4^{n - 5}$

Solution

A = strings that start with $ ad $.

A = first 2 bits are set in stone while other $ n-2 $ bits can be a, b, c, or d.

$ |A| = 1 \cdot 4^{n-2} $

B = strings that end with $ dcb $.

B = last 3 bits are set in stone while the final $ n-3 $ bits can be a, b, c, or d.

$ |B| = 1 \cdot 4^{n-3} $

$ A \cap B $ = strings that start with $ ad $ and end with $ dcb $.

Assuming the string starts with $ ad $ and ends with $ dcb $, there are $ n-5 $ bits left that can be a, b, c, or d.

$ |A \cap B| = 1 \cdot 4^{n-5} $

$ |A \cap B| = 1 $

$ |A \cup B| = |A| + |B| - |A \cap B| $

$ |A \cup B| = 1 \cdot 4^{n-2} + 1 \cdot 4^{n-3} - 4^{n-5} $

$ |A \cup B| = 4^{n-2} + 4^{n-3} - 4^{n-5} $