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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.
2. A permutation is a way of rearranging items, a combination is a way of combining different items together. What you want is to combine one item from each set (where the different sets are the set of shirts, set of pants etc') I have added links to Wikipedia for permutations and combinations, it may be worth while reading them.
Permutation with Repetition is the simplest of them all: 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)
You are looking for references on combinatorics. There are many lecture notes around the 'net, hiding under names like "discrete mathematics" (elementary combinatorics of the sort you are looking for is one of the main dishes in such a course).
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.
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 ...
Here's another way to think about it. If everyone shook hands with everyone you'd have 17∗16 2 = 136 17 ∗ 16 2 = 136 handshakes (divide by 2 because the above is double counting A shaking hands with B and so forth). However, there are 17∗2 2 = 17 17 ∗ 2 2 = 17 handshakes that aren't happening because neighbors aren't shaking hands (2 ...
Sorry This is not a complete answer but i hope it can help you . Rule of sum:" if we have (a) ways of doing something and (b) ways of doing another thing and we can not do both at the same time, then there are a + b ways to choose one of the actions."
3. Your method both under and over counts. It undercounts by neglecting patterns like GGBBGB G G B B G B, it overcounts by both permuting and then repopulating patterns. If, instead, you skip the permutations (which can all be realized through the later repopulation) you would get 23 × 3! × 3! = 288 2 3 × 3! × 3! = 288 which is the correct ...
So to count this we will use combinations (because order does not matter). And so you need to find the number of ways in which you can choose two of the six spots, which is where the H s will go. Combinations, because the order in which the H s are placed does not matter: deciding the first and then the third tosses will be head amounts to the ...