Algebra II: Computing combinations and permutations
This is a free lesson from our course in Algebra II
This lesson introduces the fundamentals and important concepts from counting principles to compute combinations and permutations. Combination and permutation are terms that refer to the selection of a subset of objects from a larger set or pool of objects. You will find some examples with solution, and the instructor explains all that with the help of audio, video presentation and in own hand writing. A permutation is a linear arrangement of a group of items. Each unique way of ordering the items of a group is known as a permutation. The number of ways to arrange a group of n items is n! (factorial of n). This means n x n-1 x n-2 x ... x1. E.g. 5! = 5x4x3x2x1= 120. The number of permutations (of length r) that can be made from a set of n items is: n! / (n-r)!, and this has the simplified notation: nPr. For Example: 6P4 = 360. (More text below video...)
<h2> Computing combinations and permutations</h2> <p> linear, equations, video, variable, example, Algebra II, substitution, system of equation, solving equation, equations, coefficients, math help, equations in three variables, practice questions, quizzes</p> <p> Combinations and permutations to compute probability</p>
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Combinations and permutations to compute probability
(Continued from above) When you have a set of objects and only want to arrange part of them, you have a permutation of n objects r at a time. For example, if you have 6 toppings for a pizza, and a customer calls and tells you to put any 3 toppings on the pizza, you might want to know how many different pizzas you can make. You can select the first topping in 6 ways, the second in 5, and the third in 4. As we learned above, this can be written as 6 * 5 * 4. so we can say the number of permutations of a set of n objects taken r at a time is given by the following formula: nPr = (n!)/(n r)!.
A combination is a set containing a certain number of objects that have been selected from another set. In combinations, the order of the elements does not matter. The number of combinations (of length r) that can be made from a set of n items is: n! / ((n-r)! r!). This has the simplified notation: nCr. E.g. 6C4 = 15. E.g. If a school has lockers with 50 numbers on each combination lock, how many possible combinations using three numbers are there. Solution: Recognize that n, or the number of objects is 50 and that r, or the number of objects taken at one time is 3. Plug those numbers in the permutation formula. 50! 50P3 = -------- (50 - 3)! = 117600.
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