Counting passwords made up of letters, digits, and special characters. Consider the following definitions for sets of characters: Digits = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} Letters = { a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z) Special characters = { *, &, $, # } Compute the number of passwords that satisfy the given constraints. (a) Strings of length 6. Characters can be special characters, digits, or letters. (b) Strings of length 7, 8, or 9. Characters can be special characters, digits, or letters. (c) Strings of length 7, 8, or 9. Characters can be special characters, digits, or letters. The first character cannot be a letter. Exercise 6.1.2: Selecting lunch specials for the week. (a) A Chinese restaurant offers 10 different lunch specials. Each weekday for one week, Fiona goes to the restaurant and selects a lunch special. How many different ways are there for her to select her lunches for the week? Note that which lunch she orders on which day matters, so the following two selections are considered different. One possible selection: Mon: Kung pao chicken Tues: Beef with broccoli Wed: Kung pao chicken Thurs: Moo shu pork Fri: Beef with broccoli A different selection: Mon: Beef with broccoli Tues: Kung pao chicken Wed: Kung pao chicken Thurs: Moo shu pork Fri: Beef with broccoli bry strings (b) Now suppose that in addition to selecting her main course, she also selects between water or tea for her drink. How many ways are there for her to select her lunches?
The Correct Answer and Explanation is :
Explanation:
(a) Passwords of Length 6:
Each character in the password can be one of the following:
- 10 digits (0-9)
- 26 letters (a-z)
- 4 special characters (*, &, $, #)
This gives a total of:
[
10 + 26 + 4 = 40 \text{ choices per character}
]
For a 6-character password:
[
40^6 = 4,096,000,000
]
Thus, there are 4.096 billion possible passwords of length 6.
(b) Passwords of Length 7, 8, or 9:
For each length, every character has 40 choices. We calculate:
[
40^7 + 40^8 + 40^9
]
[
40^7 = 16,384,000,000
]
[
40^8 = 655,360,000,000
]
[
40^9 = 26,214,400,000,000
]
Summing them:
[
16,384,000,000 + 655,360,000,000 + 26,214,400,000,000 = 268,861,440,000,000
]
Thus, there are 268.86 trillion possible passwords of length 7, 8, or 9.
(c) Passwords of Length 7, 8, or 9 (Restricted First Character):
The first character cannot be a letter, so we only have:
- 10 digits
- 4 special characters
Total = 14 choices for the first character.
The remaining characters (positions 2 onward) still have 40 choices.
[
14 \times 40^6 + 14 \times 40^7 + 14 \times 40^8
]
[
14 \times 4,096,000,000 = 57,344,000,000
]
[
14 \times 163,840,000,000 = 2,293,760,000,000
]
[
14 \times 6,553,600,000,000 = 91,750,400,000,000
]
Summing them:
[
57,344,000,000 + 2,293,760,000,000 + 91,750,400,000,000 = 94,101,504,000,000
]
Thus, there are 94.10 trillion possible passwords when the first character is restricted.
Lunch Specials:
(a) Without Drink:
Fiona has 10 lunch choices per day for 5 days. Since order matters, we use:
[
10^5 = 100,000
]
So, there are 100,000 different ways for Fiona to pick her lunches.
(b) With Drink:
Each day, she picks one of 10 meals and one of 2 drinks (water or tea). This means she has:
[
10 \times 2 = 20 \text{ choices per day}
]
For 5 days:
[
(10 \times 2)^5 = 20^5 = 3,200,000
]
Thus, there are 3.2 million different ways for Fiona to select her lunch with a drink option.
Final Answers:
- Passwords of length 6: 4.096 billion
- Passwords of length 7, 8, or 9: 268.86 trillion
- Passwords of length 7, 8, or 9 (first character restricted): 94.10 trillion
- Lunch choices without drink: 100,000
- Lunch choices with drink: 3.2 million