Solution

By product rule, there are $18 \times 325 = 5850$ ways to pick up such two representatives.

Solution

By sum rule, there are $18 + 325 = 343$ ways to pick up such representative.

Solution

By product rule, there are $4 \times 6 = 24$ routes from Boston to LA via Detroitt.

Solution

For each bit, there are 2 choices ($0$ or $1$). By product rule, there are $2^8$ of such bit strings.

Solution

Let the length of such bit string be $n | 0 < n \leqslant 6$.
1. For each bit, there are $2$ choices($0$ or $1$). By product rule, there are $2^n$ of such bit strings.
2. when $n=6$, there are $2^6$ of such bit strings; $n=5$, there are $2^5$ of such bit strings, etcetera.
3. By sum rule, there are total $2^6 + 2^5 + 2^4 + 2^3 + 2^2 + 2^1$ of such bit strings.

Solution

1. The number of strings of five ASCII characters is $128^5$.
2. The number of strings of five ASCII characters that do not contain @ is $127^5$.
3. By exclusion, the number of strings of five ASCII characters contain @ at least once is $128^5-127^5$.

Solution

1. When plates use three digits followd by three uppcase English letters, each of the first three characters has $10$ choices, and each of the last three characters has $26$ choices. By product rule, there are total $10^3 \cdot 26^3$ ways of combinations.
2. When plates use three uppercase English letters followd by three digits, each of the first three characters has $26$ choices, and each of the last three characters has $10$ choices. By product rule, there are total $26^3 \cdot 10^3$ ways of combinations.
3. By sum rule, there are total $10^3 \cdot 26^3 + 26^3 \cdot 10^3 = 2 \cdot 10^3 \cdot 26^3$ ways of combinations.

Solution

a. There are $26$ choices for each character. By product rule, there are total $26^8$ such strings.

Solution

There are $26$ choices for the first character; there are $26-1=25$ choices for the 2nd character; there are $26-2=24$ choices for the 3rd character, ..., and there are $26-7=19$ for the 8th character. By product rule, there are total $26 \times 25 \times 24 \times 23 \times 22 \times 21 \times 20 \times 19$ such strings.

Solution

There are only $1$ choice for the first character, and there are $26$ choices for each of the rest characters. By product rule, there are total $1 \times 26^7 = 26^7$ such strings.

Solution

There are only $1$ choice for the 1st character; there are $26-1=25$ choices for the 2nd character; there are $26-2=24$ choices for the 3rd character,..., there are $26-7=19$ choices for the 8th character. By product rule, there are total $1 \times 25 \times 24 \times 23\times 22\times 21 \times 20\times 19$ such strings.

Solution

There are $1$ choice for both the lst and last characters, and there are $26$ choices for each of the rest 6 characters. By product rule, there are total $1 \times 26^6 \times 1$ such strings.

Solution

There are $1$ choice for both the 1st and 2nd characters, and there are $26$ choices for each of the rest 6 characters. By product rule, there are total $1 \times 1 \times 26^6$ such strings.

Solution

Let the length of set I be $m=10$, and the length of set II be $n$ ($n<m$ in this case), and the total number of different functions from set I to set II is $C$.
There are $n$ ways to map each element of set I to set II. By production rule, there are total $n^m$ ways of such mappings, namely $C=n^m=n^{10}$.
a. $C|_{n=2} = 2^{10}$   b. $C|_{n=3} = 3^{10}$  c. $C|_{n=4} = 4^{10}$   d. $C|_{n=5} = 5^{10}$

Solution

Let the 1-to-1 function be $f$
case 1: if $n>2$, there is no 1-to-1 functions.
case 2: if $n=2$, there are $2$ 1-to-1 functions. These functions are

$\begin{aligned} f_1(1) &= 0, f_1(2) = 1 \\ f_2(1) &= 1, f_2(2) = 0 \end{aligned}$

case 3: if $n=1$, there are $2$ 1-to-1 functions. These functions are

$\begin{aligned} f_1(1) &= 0\\ f_2(1) &= 1 \end{aligned}$

Solution

There are $2$ ways to map each element , that is not $1$ nor $n$, in set I, to set II. There are only $1$ way to map element $1$ and $n$ to set II ($f(1) = 0, f(n)=0$). By production rule, there are total $1\times 2^{n-1} \times 1$ functions from set I to set II.

Solution

1. The number of bit strings that begin with two 0s is $1^2 \times 2^5 = 2^5$.
2. The number of bit strings that end with three 1s is $2^4 \times 1^3 = 2^4$.
3. The number of bit strings that begin with two 0s and end with three 1s is $1^2 \times 2^2 \times 1^3 = 2^2$.
4. By inclusion and exclusion, the number of such string is $2^5+2^4-2^2=44$.

Solution

1. The number of bit strings that begin with three 0s is $1^3 \times 2^7 = 2^7$.
2. The number of bit strings that end with two 1s is $2^8 \times 1^2 = 2^8$.
3. The number of bit strings that begin with three 0s and end with two 1s is $1^3 \times 2^5 \times 1^2 = 2^5$.
4. By inclusion and exclusion, the number of such string is $2^7+2^8-2^5=352$.

Solution

Let the length of the password be an integer $n \mid 8 \leqslant n \leqslant 12$
There are $26+26+10+6=68$ choices for each character. By product rule, the number of different passwords with length of $n$ is $68^n$. By sum rule, the total number of such passwords is $68^8 + 68^9 + 68^{10} + 68^{11} + 68^{12}$.

Solution

Let the total number of passwords in part a) be $C_{total}$, the number of password that does not contain any of the six special characters be $C_{nospecial}$, and the number of passwords contain at least one occurence of at least one of the six special characters be $C_{special}$.
1. By exclusion, $C_{special} = C_{total} - C_{nospecial}$.
2. According to part a), $C_{total} = 68^8 + 68^9 + 68^{10} + 68^{11} + 68^{12}$. Similarly, $C_{nospecial} = 62^8 + 62^9 + 62^{10} + 62^{11} + 62^{12}$.
3. $C_{special} = 68^8 + 68^9 + 68^{10} + 68^{11} + 68^{12} - (62^8 + 62^9 + 62^{10} + 62^{11} + 62^{12})$

Solution

The time is $(68^8 + 68^9 + 68^{10} + 68^{11} + 68^{12}) \times 10^{-9}s \approx 314582 \: years$

Solution

Let the length of the variable name be an positive integer $n |0 < n\leqslant 8$.
1. There are $26+26+1 = 53$ choices for the 1st character, and there are $26+26+10+1 = 63$ choices for the rest of each character.
2. By product rule, there are total $53 \times \overbrace{63 \times 63 \cdots \times 63}^{n-1 \: times} = 53 \times 63^{n-1}$ such different variables with length of $n$.
3. By sum rule, there are total $53 \times 63^{8-1} + 53 \times 63^{7-1} + \cdots \times 53 \times 63^{1-1} = 53(63^7 +63^6+ \cdots + 63^0)$ different variables can be named in C.