jimmycricket said:Given a list (a,b,c,d,e,f) What method can I use to quickly find all 15 pairings of two letters. I can do it without using a specific decision procedure throughout but sometimes combinations get repeated and its hard to check quickly which ones have already been. Essentially I would like an algorithm to find every combination of pairs in a set of arbitrary length/
jimmycricket said:i don't follow that. Please note I don't know programming i need a more wordy answer please
There are 15 ways to pick pairs of objects from a list of 6. This can be calculated using the combination formula nCr = n!/(r!(n-r)!), where n is the total number of objects and r is the number of objects being chosen at a time. In this case, n = 6 and r = 2, so the formula becomes 6C2 = 6!/(2!(6-2)!) = 15.
The probability of picking two specific objects from a list of 6 depends on the total number of objects in the list and the number of objects being chosen at a time. In this case, the probability can be calculated using the formula P = (1/n)(1/(n-1)), where n is the total number of objects. So for a list of 6 objects, the probability would be (1/6)(1/5) = 1/30.
Yes, you can pick more than two objects at a time from a list of 6. The number of objects that can be chosen at a time is called the sample size. In this case, the sample size is 2, but it can be increased or decreased depending on the specific scenario.
The number of objects in the list directly affects the number of possible pairs that can be chosen. The larger the number of objects, the more possible combinations there are. For example, if the list had 10 objects instead of 6, there would be 45 possible pairs instead of 15.
Picking pairs with replacement means that an object can be chosen more than once, while picking pairs without replacement means that an object can only be chosen once. In the case of picking pairs from a list of 6 objects, the number of possible pairs with replacement would be 36, while the number of possible pairs without replacement would be 15.