Calculating 4-Character Password Possibilities
The problem asks for the number of distinct 4-character passwords that can be formed using digits $0 - 9$ and letters $A - F$ without repetition.
Character Set Identification
- Digits available: $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ (10 digits)
- Letters available: $A, B, C, D, E, F$ (6 letters)
- Total available unique characters: $10 + 6 = 16$
- Password length required: 4 characters
- Constraint: No character repetition allowed
Permutation Calculation
Since the order of characters matters in a password and repetition is not allowed, this is a permutation problem. We need to find the number of permutations of selecting 4 characters from a set of 16 distinct characters.
The formula for permutations is $P(n, k) = \frac{n!}{(n-k)!}$, where $n$ is the total number of items to choose from, and $k$ is the number of items to choose.
In this case, $n = 16$ and $k = 4$.
Number of possible passwords = $P(16, 4)$
$ P(16, 4) = \frac{16!}{(16-4)!} = \frac{16!}{12!} $ $ P(16, 4) = 16 \times 15 \times 14 \times 13 $Result
Performing the multiplication:
- $16 \times 15 = 240$
- $14 \times 13 = 182$
- $240 \times 182 = 43680$
Therefore, there are 43,680 possible 4-character passwords.
Important Questions from Permutation and Combination (Notes)
- In how many ways can 10 men be divided into two groups of 4 men and 6 men?
- How many 5-digit numbers can be formed from the digits 0, 2, 3, 4, 6, 7 and 9, using each at most once, which are divisible by 5?
- In how many ways can you place $N$ coins on a board with $N$ rows and $N$ columns such that every row and every column contains exactly one coin?
- From a group of 40 players, a cricket team of 11 players is chosen. Then, one of the eleven is chosen as the captain of the team. The total number of ways this can be done is
[$\binom{m}{n}$ below means the number of ways $n$ objects can be chosen from $m$ objects] - The maximum number of points formed by intersection of all pairs of diagonals of convex octagon is
