![]() ![]() Remember that the order in which I pick them doesn't make any difference. Now I want to play the game in teams of 3 and need to select 2 of my 4 friends to form my team. This gives us our 3 words: OPT, TOP and POT. We can find the different words which can be created from the set of three letters using permutation and filter out the words which have a meaning. For example, in our problem, we can arrange the three letters in the following 6 ways. Permutation refers to the different ways in which a given set of objects can be arranged. We can find all the words using the different arrangements of the four letters. We can use brute force to arrange the letters in the word OTP in all possible positions. (Hint: There are 3 words) Well, I am able to guess only two: POT and TOP :disappointed: How to find out the third? Let's play a game :smiley: Try to form as many words as you can by using all the letters: O, T, P. We can find the permutations and the combinations of a word or a set of numbers using recursion as well as pre-defined methods in the Python library itertools. Combinations are the ways in which we can select a certain subset of items from a bigger list, irrespective of the order of selection. Result_list.Permutations refer to the different ways in which we can arrange a given list of elements. # and add the permutations to the result list # by the old prefix extended by element el # recursively create the rest of the permutations prefixed # the prefix is a full permutation -> nothing to permute left If you want to create a recursive function without using itertools, you could do it as follows: def create_permutations(base_list, prefix_list): The result would be: ('Orange', 'Apple', 'Banana', 'Peach', 'Avocado') Rest_elements= fruits + fruitsįor permutation in permutations(rest_elements): # create a temporary list with all elements that still are You could just remove the fixed element, then create the permutation over the rest and add the fixed to the result, like this: from itertools import permutationsįruits = Permutation(, list(range(1, len(fruits)))) ![]() #this is alternative to the one above NOT in the same function #can only concatenate list (not "int") to list #the second option which gives TypeError: #the first try which gives the permutations Here I am asking how this particular function would need to be corrected? fruits = NOTE: I can get it done with itertools.permutations OR with permutations with all the elements OR with doing it first with four elements and then adding the first one when printing What is wrong and what is needed to fix the code. I have tried to solve with two alternative ways, however, the first one keeps adding elements to the same result list, and the other one results a typeError. I cannot find answers to this specific permutation problem, thus, I am asking it with a new question. I need a print with a certain format after each permutation (Fruit1 Fruit2 Fruit3 Fruit4 Fruit5) with the first element being the same all the time. I am doing an exercise on recursive permutations with a fixed first element. ![]()
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