![]() ![]() A permutation is called a subset that can be provided in the calculation. Often you can just say “lexicographic ordering” and you're done.The permutation is the part of Mathematics that can be allowed where repetition belongs. Sometimes you just need to know that some objects can be ordered so a proof/algorithm works out. Aside: lexicographic ordering comes up in weird places.For example, if we are permuting the letters \(ABCDE\), then the permutation \(ABDEC\) is less than \(ABEDC\).A permutation \(A\) is greater than \(B\) if \(B\) is less than \(A\).Two permutations are equal if all elements are the same. ![]() We will say that a permutation \(a_1a_2\ldots a_n\) is less than \(b_1b_2\ldots b_n\) if either:.… which is a fancy way to say “dictionary order”.We will order them based on lexicographic order.Then we can generate them in that order, and be sure we found them all. We should first decide on a way to order permutations.you want to write an automated test for code that should be able to handle input in any order. you want to do write code to search for a perm/comb that satisfies some condition. It's occasionally useful to actually generate all of the permutations/combinations, not just count them.Then distribute the remaining 925 in \(C(949,25)\) ways. How many of those are there? Ummmm…Īttempt #2: First give everybody three pens, using 75 of them. Attempt #1: Count all \(C(1024,25)\) ways to distribute the pens, then subtract the number where somebody gets 2 or fewer pens.So there are \(C(1024,25)\) ways to distribute. This is a combination with repetition problem: combinations of 1000 the 25 family members with repetition. Order doesn't matter here, since the pens are identical. To distribute the gifts, we must select people to get each one. See the textbook's discussion of “distinguishable objects and indistinguishable boxes” on p. 337, or look up Stirling Numbers of the second kind. Example: How many ways are there to permute the letters of the word \(\mathrm\.\].The basic approach is: how can you make it look like a problem you know how to solve? When you do, how much did over-/undercount by?.There are many problems similar to the basic combination/permutation ones.In summary we have these ways to select \(r\) things from \(n\) possibilities:.We select 100 of them (with repetition) and that gives us a solution to the equation. In other words, we have balls labelled \(a\), \(b\), and \(c\).In order to satisfy the equation, we have to select 100 “ones”, some that will contribute to \(a\), some to \(b\), some to \(c\).Example: How many solutions does this equation have in the non-negative integers?.Proof: like with the candy, but not specific to \(r=6\) and \(n=3\).If we are selecting an \(r\)-combination from \(n\) elements with repetition, there are \(C(n+r-1,r)=C(n+r-1,n-1)\) ways to do so.… or equivalently, there are \(C(8,6)=28\) ways to place the candy selections.There are \(C(8,2)\) ways to do that, so \(C(8,2)=28\) possible selections.Now the answer becomes obvious: we have 8 slots there and just have to decide where to put the two dividers.Now we don't need the actual identities in the diagram to know what's there:.We don't want our candy to mix: let's separate the types.Since order doesn't matter, we'll list all of our selections in the same order: C then G then H.Here are some possible selections you might make:.How many different selections can you make? The store has chocolate (C), gummies (G), and Haribo sugar-free gummi-bears, which are horrible (H). Example: You walk into a candy store and have enough money for 6 pieces of candy.Same as permutations with repetition: we can select the same thing multiple times.Same as other combinations: order doesn't matter.We can also have an \(r\)-combination of \(n\) items with repetition.Proof: the product rule applied \(r\) times.There are \(n^r\) different \(r\)-permutations of \(n\) items with repetition.Specifically, we select \(r\) objects from \(n\) possibilities, and are allowed to select the same object as many times as we want.That was an \(r\)-permutation of \(n\) items with repetition allowed.We can actually answer this with just the product rule: \(10^5\).Here we are selecting items (digits) where repetition is allowed: we can select 4 multiple times if we want.Example: How many distinct values can be represented with 5 decimal digits?.Something in particular that was missing: being able to select with repetition.… and we found problems where those were useful, but it wasn't obvious.In how many ways if order does/doesn't matter? You have \(n\) objects and select \(r\) of them.The permutation and combination question we have done so far are basically about selecting objects.
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