For each of these 30 possibilities of seating these first two people, there'd be four possibilities of who we put in chair number three. Of who we'd put here, because one person's already sitting down. For each of those six possibilities there would be five possibilities We seat them in order, we might as well, weĬan say well there'd be six possibilities here. People into four chairs? Well the first chair, if Permutations are there of putting these six We can go through this fairly quickly, one, two, three, four chairs. Person A, B, C, D, E, and F, so we have six people. Start using other examples, other than just people sitting in chairs, but let's just stick with it for now. But each incrementalĮxample I'm gonna review what we've done before, but hopefully go a little bit further. So I think it never hurts to do as many examples. I've been struggling for over a week with this.Įxposed to permutations and combinations it takes a little bit to get your brain around it, Leaving us with the FIRST PERMUTATION that was counted or ONE COMBINATION of those 4 letters. Essentially, what the k! is doing by dividing the TOTAL PERMUTATIONS, is it is canceling all the additional permutations that were counted MORE THAN ONCE in the permutations formula. But were not counting permutations only COMBINATIONS, thus all we want to count is the FIRST PERMUTATION of the four letters. That means ABCD is 1 COMBINATION but it has 4! PERMUTATIONS (ABCD, ADCB, DCBA.etc). In light of this, it becomes clear that the permutation formula is counting one combination k! times, k = the number of chairs or spots (4 in the video). This reminded me of the principle of Inclusion/Exclusion which is basically about subtracting over counted elements. It's deceiving because the k! is actually DIVIDING the entire permutation equation. The part that confused me about the combinations formula is what the multiplication of k! in the denominator is doing to the formula. the number of permutations is equal to n!/(n-k)! so the number of combinations is equal to (n!/(n-k)!)/k! which is the same thing as n!/(k!*(n-k)!). So the formula for calculating the number of combinations is the number of permutations/k!. The group size can be calculated by permuting over the number of chairs which is equal to the factorial of the number of chairs(k!). So the number of combinations is equal to the number of permutations divided by the size of the groups(which in this case is 6). If we didn't care about these specific orders and only cared that they were on the chairs then we could group these people as one combination. So some of the permutations would be ABC, ACB, BAC, BCA, CAB and CBA. In our example, let the 5 people be A, B, C, D, and E. The number of combinations is the number of ways to arrange the people on the chairs when the order does not matter. So the formula for the number of permutations is n!/((n-k)!. For n people sitting on k chairs, the number of possibilities is equal to n*(n-1)*(n-2)*.1 divided by the number of extra ways if we had enough people per chair. We can make a general formula based on this logic. So the total number of permutations of people that can sit on the chair is 5*(5-1)*(5-2)=5*4*3=60. On the third chair (5-2) people can sit on the chair. On the second chair (5-1) people can sit on the chair. If there are 3 chairs and 5 people, how many permutations are there? Well, for the first chair, 5 people can sit on it.
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