![]() ![]() 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. Hope this helps out and don't hesitate to reach out if it doesn't.Ok, let's start by an example. Essentially, as long as it matters who we put where, we have variations. The same distinction can be assigned to the tennis example, where we can name the positions: "winner", "runner-up" and "not in final". For example, when we have to match banners to social media platforms in question 2, we have this artificial "order" because every position (platform) is different. Thus, when some (or all) position matter, we are dealing with variations. Then, we have P(3,2) = 3! / 1! = 6 ways they 3 competitors can arrange. Now, if we care who lifts the trophy, we use variations because order is relevant. The 3 combinations are, obviously, Djokovic vs Nadal, Nadel vs Federer or Djokovic vs Federer. Hence, if you only care about the match up, but don't care who actually ends up as the victor, you use combinatorics -> C(3,2) = 3!/(2!*1!) = 3. If you simply care which 2 made the final, but not who won, we would use combinations because order does not matter. You know all 3 men were in the tournament and 2 of them reached the final. So, the main distinction between the two is that combinations don't care about order, while variations do.įor instance, suppose you love tennis and you're a big fan of Djokovic, Nadal and Federer. You can think of them as a special case of Variations where n = p and just distinguish between using variations and combinations. As for the difference between the two, let's start with the difference between the two and work through a simple example. So, Permutations are when we arrange the entire set. ![]()
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