A robber has stolen a locked briefcase. The lock on the briefcase can be opened with a three-digit code, where each digit is one of the digits from 0 to 5.a) How many codes are possible if each digit may be used more than once?b) How many codes are possible if each digit may be used only once?
Please show more than one approach.
a) Is it 6.6.6?
b) is it 6.5.4?
agree
However, online, “*” is used to indicate multiplication to avoid confusion with decimals.
a) To find the number of possible codes if each digit may be used more than once, we need to consider the number of choices for each digit. Since each digit can be any number from 0 to 5 (inclusive), there are 6 choices for each digit. Since there are 3 digits in the code, we can use the multiplication principle to determine the total number of possible codes:
6 choices for the first digit x 6 choices for the second digit x 6 choices for the third digit = 6^3 = 216 possible codes
Alternatively, we can also approach this problem using combinatorics. Since each digit has 6 choices, we can think of this as a combination with repetition problem. We need to select 3 digits from a set of 6 possibilities, allowing for repetition. The number of such combinations can be calculated using the formula:
n + r - 1 choose r
where n is the number of choices and r is the number of selections. In this case, n = 6 (choices for each digit) and r = 3 (number of digits in the code). Plugging these values into the formula, we get:
6 + 3 - 1 choose 3 = 8 choose 3 = (8 * 7 * 6) / (3 * 2 * 1) = 56 possible codes
b) To find the number of possible codes if each digit may be used only once, we need to consider the number of choices for each digit, taking into account that once a digit is chosen, it cannot be reused. For the first digit, we have 6 choices as before. However, for the second digit, we have only 5 choices remaining (since one digit has already been chosen). Similarly, for the third digit, we have only 4 choices remaining (since two digits have already been chosen). We can use the multiplication principle to determine the total number of possible codes:
6 choices for the first digit x 5 choices for the second digit x 4 choices for the third digit = 6 * 5 * 4 = 120 possible codes
Alternatively, we can approach this problem using combinatorics. We can think of this as a permutation problem, where the order of selection matters. We need to select 3 digits from a set of 6 possibilities without repetition. The number of such permutations can be calculated using the formula:
n! / (n - r)!
where n is the number of choices and r is the number of selections. In this case, n = 6 (choices for each digit) and r = 3 (number of digits in the code). Plugging these values into the formula, we get:
6! / (6 - 3)! = 6! / 3! = 6 * 5 * 4 = 120 possible codes