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3. There is some common confusion between these two terms. The word "permutation" in general refers to one of three things depending on context. It can mean the order (arrangement) of a set as in combinatorics. Or it can refer to an arrangement of a subset of a given size as also in combinatorics.
For this you would want to use combinations not permutations, since the dice are indistinguishable. $6^3$ gives the number of possible rolls under the assumption that $(3,3,2)$ is different than $(3,2,3)$ but since this is a casino and casinos don't have numbered or different colored die, $(3,3,2)$ should be the same as $(3,2,3)$ since there is no way to tell these two results are distinct.
Permutation formula for ordered with repetition is n^r where n is the number of things to choose from and r is how many we are choosing to form another set. Total possible permutations is 62^8 However the rules state one numeric and one alpha must be used. The largest legal set considering all rules is 52^7 + 10^1.
Through some browsing I've found that the number of combinations with replacement of n items taken k at a time can be expressed as ((n k)) [this "double" set of parentheses is the notation developed by Richard Stanley to convey the idea of combinations with replacement]. Alternatively, ((n k)) = (n + k − 1 k). This is more familiar notation.
N to the power of R. Example: 3 tosses of 2-sided coin is 2 to power of 3 or 8 Permutations possible. In these, "at-least-2 Heads in a row" permutations are: HHH, HHT, THH - 3. Probability of "at least 2 heads in a row" is 3/8th (0.375) If the question is "If you throw a 2-sided coin (N=2), R times, how many times can you get at least 2 heads ...
But there are C(n, k) combinations. So P(n, k), defined as the total number of permutations of k items selected from n distinct items, obeys the equation P(n, k) = C(n, k) × k!. Now it is a simple arithmetic operation to divide by k! on both sides, with the result C(n, k) = P(n, k) k!. And that's why we divide the number of permutations by k!.
So this is a case pf permutations but where certain outcomes are equal to each other. Hence the total combinations of r picks from n items is n!/r!(n-r)! I haven't discussed the mathematics of deriving the equation in depth.
Specifically, the part where we divide the number of permutations by the factorial of the repetition. From the textbook : In general, repetitions are taken care of by dividing the permutation by the factorial of the number of objects that are identical.
There are 1 × 2 × 2 = 4 1 × 2 × 2 = 4 possible selections that can be made. For each selection, there are 3! = 6 3! = 6 ways to permute the objects, giving a total of 4 × 6 = 24 4 × 6 = 24 permutations for three objects. Altogether, this is a grand total of 1 + 5 + 16 + 24 = 46 1 + 5 + 16 + 24 = 46 permutations. Share.
2. Permutations are used when one is concerned with order. For example, if you wanted to choose how many ways are there to arrange five people in a line, the answer will be different. In your case, a person first in line is the same as a person fourth in line, i.e., order does not matter. So combinations are used when the problem does not ...