$$3\mapsto 4\ \implies f\circ g = (12)(34\dotsc\\Ĥ\mapsto 2 \to 3 \implies f\circ g = (12)(34)\dotsb\\ĥ\mapsto 3\to 5 \implies f\circ g = (12)(34)(5)$$Įvaluate all possible combinations of $f$, $g$, and $h$, and you will find the permutation your looking for. Evaluating from right to left, starting at the 3rd cycle, it first appears in the 2nd cycle, where it's mapped to 3, so $2\mapsto 5\mapsto 3$, looking for the three and finding it in the first cycle, you'll get $2\mapsto 5\mapsto 3\mapsto 1$, hence In the fourth line below we see how the expression can be simplified using factorial notation. In the second line below, we have multiplied by, which is just the value 1 because the numerator and denominator are equal. The idea is like factoring an integer into a product of primes in this case, the elementary pieces are called cycles. We can represent permutations more concisely using cycle notation. So you have to check where $5$ is getting mapped to. There is a step of arithmetic we can apply to the general pattern for P(n,r) to help streamline permutation calculations. Permutation notation is ne for computations, but is cumbersome for writing permutations. It appears first in the 4th first cycle, where it's getting mapped to 5, so $2\mapsto 5$. To simplify this, we use factorial notations. In plain language, this represents the total number of permutations from choosing 2 items from a collection of 4 items. The word 'permutation' also refers to the act or process of changing the linear order of an ordered set. So you have to check, where the $2$ is going. The letter P in the nPr formula stands for permutation which means arrangement. In mathematics, a permutation of a set is, loosely speaking, an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements. Which in this case this is again $1$, which (evaluating from right to left, starting at the second cycle, since you've already used the 4th and the 3rd) next appears in the first cycle, where it's getting mapped to $1$, so $1\mapsto 1\mapsto 2$. Now you have to check, where the target is going. In particular, you have to evaluate from right to left, since the right map is applied first.įor example, if you want to evaluate $f\circ g$ in your exercise, first write down the composition of cycles:Īnd then look where every element's mapped to: evaluating from right to left and chosing $1$ as our starting point, it first appears in the 3rd cycle, which yields $1\mapsto 1$. To evaluate this, you have to evaluate the maps and look where every element is mapped to. So composition is nothing else than composition of maps. Note that $\sigma = (123)(4)(5)$ is just another notation for the map $$\sigma\colon\\\
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