If it’s a permutation, just do the numbers on top of the slots if it’s a combination, you have to do the numbers on top and the numbers on the bottom. The number of groups of dogs that Brandon can take on his walk is: ![]() Draw three slots, start with the total number of choices, and count down:īut now on the bottom of the underscores, start with the number 1 and count up: Begin by doing exactly what you did in the permutation example. But don’t worry, there’s a simple method once again. The only way to find the number of Combinations is by first finding the number of permutations, then dividing by the number of ways each group can be arranged. Then count down as you fill each slot:Ĭombination questions, such as the one with Brandon and his dogs, are a bit more complicated. If the question is a permutation, such as the Ugliest Dog example, you start by filling in the total number of items you have to choose from (in this case, 8,) in the first slot. For example, in both of the above problems, we’re selecting 3 dogs, so we draw three slots: First, draw a series of underscores, one for each person or item you’re selecting. The method for doing these problems is simple. So if you see the word groups, you’re probably dealing with a combination question. It’s still the same group of dogs going on the walk. In other words, if he calls dog A, then dog B, then dog C, or dog C, then dog B, then dog A, it doesn’t matter. This is a combination question because once Brandon chooses the three dogs, the order in which he arranges them doesn’t matter. If Brandon owns 8 dogs, how many different groups of dogs can he choose for his walk? In fact, if you see the word arrangements in the question, you’re always dealing with a permutation.īrandon wants to take 3 dogs with him on his morning walk. These are two distinct arrangements, or permutations. In other words, A-B-C and C-B-A are considered two different outcomes. In other words, if the eight dogs are A,B,C,D,E,F,G, and H, and the ribbons go to dogs A,B, and C, it matters whether dog A is first, B is second, and C is third, or C is first, B is second, and A is third. ![]() Why? Because not only are we choosing three dogs, but the order in which they are arranged makes a difference. If there are 8 dogs in the competition, how many different ways can the ribbons be awarded? In the Ugliest Dog competition, a blue ribbon will be awarded to the ugliest dog, a red ribbon to the second ugliest dog, and a yellow ribbon to the third ugliest dog. Combinations ask how many different groups of people or items can be chosen from a larger group. ![]() Permutations how many different arrangements can be created from a group of people or items. Don’t worry – there’s an easier way.įirst let’s go over the difference between Permutations and Combinations. With $3$ people and $3$ awards, there would be $3^3=27$ ways to give these awards to the people.If you’ve been studying Permutations and Combinations, you’ve probably had to memorize a bunch of ugly formulas involving factorials. ![]() If we can give multiple awards to each person, we have replacement. Applied to your problem, imagine a variation in which instead of ranks we were giving $A,B,$ and $C$ awards. But since we put it back, on the next draw we also have $3$ possibilities! Thus, if we draw $k$ times, we have $3^k$ ways to pull the names. For the first name we pull, we have $3$ possible names to pull. How many ways are there to pull the names? Say we have $3$ names. This occurs when you allow replacement of the objects! Imagine you're pulling names from a hat, but every time you pull a name and read it, you put it back. The $3^3=27$ answer comes in a slightly different variation which involves neither permutations or combinations. To arrange $n$ items in a row (which can be accomplished in $n!$ ways) is equivalent to picking $k$ of $n$ items to arrange in a row (which can be accomplished in $\frac=3!$ - exactly the result we were looking for!
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