If we had only one character repeated, the problem is finished, and the final result would be TOTAL - INVALID permutations.īut, if we have more than one repeated character, we have counted some of the invalid strings twice or more times. Of course, as we also have two 'f', the number of permutations with 'ff' will be the same as the ones with 'aa': So, the total number of permutations with two 'a' together is: We also have to consider that the two 'a' can be themselves permuted in 2! (as we have two 'a') ways. Now we have only six boxes, as one of them will be filled with 'aa': Our example string has two repeated characters: the 'a' appears twice, and the 'f' also appears twice. As they have to be together, why don't consider them something like a "compound" character? To calculate them, we'll consider that any character that has the same character side by side, will be in the same box. So the total number of permutations, without restrictions, is:Īny permutation with two equal characters side by side is not valid. We can fill the first with any of the 7 characters, the second with any of the remaining 6, and so on. Which has 7 characters, there are seven boxes to fill. Let's consider each position a small box. We have to fill a number of positions, that is equal to the length of the original string. To find the solution we have to calculate the total number of permutations (without restrictions), and then subtract the invalid ones. This is a mathematical approach, that doesn't need to check all the possible strings. The maths to resolve the case where only one character is repeated is pretty trivial - Factorial of total number of characters minus number of spaces available multiplied by repeated characters.īut trying to figure out the formula for the instances where more than one character is repeated has eluded me. I posted this question on math.stackexchange. But I really want to try to resolve this problem efficiently rather than brute force through an array of all possible permutations. I have read through a lot of post about Combinatorics and Permutations and just seem to be digging a deeper hole for myself. The tests for this problem are as follows (returning the number of permutations that meet the criteria)- "aab" should return 2 The problem sees each character as unique so maybe "aab" could better be described as "a1a2b" The example given in the problem is "aab" which the site says has six possible permutations, with only two meeting the non-repeated character criteria: Second question - If yes, what formula could I use? I know I can solve this by writing a program that creates every permutation and then filters out the ones with repeated characters.īut I have this gnawing feeling that I can solve this mathematically. Have the same letter (in this case 'a') repeating. Return 2 because it has 6 total permutations, but only 2 of them don't Return the number of total permutations of the provided string thatĭon't have repeated consecutive letters.
Number of permutations code#
Hence the multiplication axiom applies, and we have the answer (4P3) (5P2).I'm working on a Free Code Camp problem. For every permutation of three math books placed in the first three slots, there are 5P2 permutations of history books that can be placed in the last two slots. So the answer can be written as (4P3) (5P2) = 480.Ĭlearly, this makes sense. Therefore, the number of permutations are \(4 \cdot 3 \cdot 2 \cdot 5 \cdot 4 = 480\).Īlternately, we can see that \(4 \cdot 3 \cdot 2\) is really same as 4P3, and \(5 \cdot 4\) is 5P2. Once that choice is made, there are 4 history books left, and therefore, 4 choices for the last slot. The fourth slot requires a history book, and has five choices. Since the math books go in the first three slots, there are 4 choices for the first slot,ģ choices for the second and 2 choices for the third. We first do the problem using the multiplication axiom. In how many ways can the books be shelved if the first three slots are filled with math books and the next two slots are filled with history books?
You have 4 math books and 5 history books to put on a shelf that has 5 slots. Since two people can be tied together 2! ways, there are 3! 2! = 12 different arrangements The multiplication axiom tells us that three people can be seated in 3! ways. Let us now do the problem using the multiplication axiom.Īfter we tie two of the people together and treat them as one person, we can say we have only three people. So altogether there are 12 different permutations.
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