Consider the following definitions for sets of characters: Digits = {0, 1, 2, 3,
ID: 3120967 • Letter: C
Question
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. Answer questions with big number answers as powers or sum of powers (Ex: 15%^7 instead of 170, 859, 375) Strings of length 6. Characters can be special characters, digits, or letters. Strings of length 7, 8, or 9. Characters can be special characters, digits, or letters. Strings of length 7, 8, or 9. Characters can be special characters, digits, or letters. The first character cannot be a letter 4) Strings of length. Strings of length 6 Characters can be special characters, digits, or letters, with no repeated characters. You will not be able to answer this with as a power or sum or powers) Strings of length 6. Characters can be special characters, digits, or letters, with no repeated characters The first character cannot be a special character.Explanation / Answer
1) There are 10 digits, 26 letters and 4 speacial characters.
So, strings of length 6?
every place could have either of the digits (10), letters (26), and special characters (4)
So, Number of passwords in this case is = 40^6 as there are total 40 (digits+letters+special characters).
2) Strings of length 7,8 or 9 = Number of passwords with strings 7 + Number of passwords with strings 8 + Number of passwords with strings 9
= 40^7 + 40^8 + 40^9
3) Strings of length 7,8 or 9 with first character cannot be a letter = Number of passwords with strings 7 + Number of passwords with strings 8 + Number of passwords with strings 9
Now first space can be filled in 40 - 26 =14 ways
So, 14*40^6 + 14*40^7 + 14*40^8 = 14 (40^6 + 40^7 + 40^8)
4) Strings of length 6 with no repetition:
First place can be filled with 40 options, 2nd place with 39 and so on
So, number of passwords = 40*39*38*37*36*35
5) Strings of length 6 with no repetition and first character cannot be a special character
First space can be filled in 40 - 4 = 36 ways
Second space = 36 + 4 - 1 = 39 as repetition is not allowed and so on
Number of passwords is thus = 36 * 39 * 38 * 37 * 36 *35