Permutation and combination :An easy method . Must read !

If you’ve been studying Permutations and Combinations, you’ve probably had to memorize a bunch of ugly formulas involving factorials. Don’t worry – there’s an easier way. Permutations vs. Combinations First let’s go over the difference between Permutations and Combinations. These two terms are easy to mix up. Permutations how many different arrangements can be created from a group of people or items. Combinations ask how many different groups of people or items can be chosen from a larger group. Here are two examples: 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. If there are 8 dogs in the competition, how many different ways can the ribbons be awarded? This is a permutation question. Why? Because not only are we choosing three dogs, but the order in which they are arranged makes a difference. 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. In other words, A-B-C and C-B-A are considered two different outcomes. These are two distinct arrangements, or permutations. In fact, if you see the word arrangements in the question, you’re always dealing with a permutation. Now consider this question: Brandon wants to take 3 dogs with him on his morning walk. If Brandon owns 8 dogs, how many different groups of dogs can he choose for his walk? This is a combination question because once Brandon chooses the three dogs, the order in which he arranges them doesn’t matter. 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. It’s still the same group of dogs going on the walk. So if you see the word groups, you’re probably dealing with a combination question. The Method The method for doing these problems is simple. First, draw a series of underscores, one for each person or item you’re selecting. For example, in both of the above problems, we’re selecting 3 dogs, so we draw three slots: ___ ___ ___ 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. Then count down as you fill each slot: 8 7 6 and multiply the numbers: 8 * 7 * 6 = 336 total permutations. Combination questions, such as the one with Brandon and his dogs, are a bit more complicated. 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. But don’t worry, there’s a simple method once again. Begin by doing exactly what you did in the permutation example. Draw three slots, start with the total number of choices, and count down: 8 7 6 But now on the bottom of the underscores, start with the number 1 and count up: {8/1}*{7/2}*{6/3} The number of groups of dogs that Brandon can take on his walk is: {8*7*6}/{1*2*3}={336/6}=56 That’s all there is to it. 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. There are just a few tips to remember: Permutations are when the order matters, combinations are when it doesn’t. If A-B-C is different from C-B-A, it’s a permutation; if A-B-C and C-B-A are the same, it’s a combination. The math is more complicated when the problem is a combination. This always confuses people: when the order doesn’t matter, the math is harder, because you have to do the numbers on the bottom. That might seem counterintuitive but that’s how it is. All of these problems assume that the same item cannot be used twice. Here are two more examples: The Gamma Zeta Beta fraternity must choose a committee of four members to plan its annual Children’s Hospital fund raiser and beer bash. If the fraternity has 10 members, how many different committees can be chosen? This is a combination question because it only matters which 4 frat boys we choose. We are not arranging them in any particular order. So the solution is: {10/1}*{9/2}*{8/3}*{7/4}={5040/24}=210 Remember, if the order doesn’t matter, you must do the numbers on the bottom! If you want to save time, reduce before you multiply: the 4 and the 2 on the bottom will cancel the 8 on top, and the 3 on the bottom reduces the 9 on top to a 3, so you have 10 * 3 * 7 = 210. The bottom can always be completely canceled out. Now try this: The Gamma Zeta Beta fraternity is electing a President, Vice President, Secretary, and Kegger Chair. If the fraternity has 10 members, in how many different ways can the officers be chosen? Now we have a permutation, because we are not only choosing 4 frat boys, but also assigning them to particular jobs. So A-B-C-D is now different from D-C-B-A. This makes our math easier: 10 * 9 * 8 * 7 = 5040.   permutation and combination for cat pdf permutation and combination cat study material permutation and combination cat tricks permutation and combination for cat preparation permutation and combination basics for cat permutation and combination concepts pdf cat probability questions with solutions