Solution: 2018 Fall Final - 4

Author: Michiel Smid

Question

Consider strings of length 70, in which each character is one of the letters $a, b, c$. How many such strings have at least 3 letters $c$?

(a)

$\sum_{k=4}^{70} {70 \choose k} \cdot 2^{k}$

(b)

$3^{70} - 2^{70} - 70 \cdot 2^{69}$

(c)

$3^{70} - 2^{70} - 70 \cdot 2^{69} - {70 \choose 2} \cdot 2^{68}$

(d)

$\sum_{k=4}^{70} {70 \choose k} \cdot 2^{70-k}$

Let S be all possible strings of length 70: $ |S| = 3^{70} $
Let A be the event that the string has no letters $c$
The number of strings with no letters $c$ is $2^{70}$
$ |A| = 2^{70} $
Let B be the event that the string has exactly 1 letter $c$
First, we choose the position of the letter $c$: $ \binom{70}{1} = 70$
Then, we choose the positions of the letters $a$ and $b$ for the remaining 69 positions: $ 2^{69} $
$ |B| = 70 \cdot 2^{69} $
Let C be the event that the string has exactly 2 letters $c$
First, we choose 2 positions out of the 70 for the letter $c$: $ \binom{70}{2} $
Then, we choose the positions of the letters $a$ and $b$ for the remaining 68 positions: $ 2^{68} $
$ |C| = \binom{70}{2} \cdot 2^{68} $

$ |A \cup B \cup C| = |S| - |A| - |B| - |C| $

$ |A \cup B \cup C| = 3^{70} - 2^{70} - 70 \cdot 2^{69} - \binom{70}{2} \cdot 2^{68} $