Permutations and combinations

Relevant definitions to solve this exercise: “THE PRODUCT RULE Suppose that a procedure can be broken down into a sequence of two tasks. If there are n1 ways to do the first task and for each of these ways of doing the first task, there are n2 ways to do the second task, then there […]

Let’s refresh the definitions. “A permutation of a set of distinct objects is an ordered arrangement of these objects. We also are interested in ordered arrangements of some of the elements of a set. An ordered arrangement of r elements of a set is called an r-permutation. ” “An r-combination of elements of a set

First, let’s examine the definitions of permutations and combinations. “A permutation of a set of distinct objects is an ordered arrangement of these objects. We also are interested in ordered arrangements of some of the elements of a set. An ordered arrangement of r elements of a set is called an r-permutation. ” “An r-combination

This is a simple exercise. To solve it, we just need to apply the formula to calculate the number of r-combinations. “The number of r-combinations of a set with n elements, where n is a nonnegative integer and r is an integer with 0 ≤ r ≤ n, equals C(n,r) = n! / r!(n−r)!” Discrete

This is another simple exercise. To solve it, we just need to apply the formula to calculate the number of 3-permutations. “If n and r are integers with 0 ≤ r ≤ n, then P(n,r) = n! / (n−r)! ”. Discrete mathematics and its applications by Rosen. Answer: We just need to calculate the number

This type of exercise is easy. You are only required to apply a formula. You just need to know the formula. “If n and r are integers with 0 ≤ r ≤ n, then P(n,r) = n! / (n−r)! ”. Discrete mathematics and its applications by Rosen. a) P(6,3) P(6,3)=6!/(6-3)! =6!/3! =6x5x4x3!/3! =120 b) P(6,5)

“A permutation of a set of distinct objects is an ordered arrangement of these objects.” We can follow a simple approach to solve this exercise. If we calculate how many permutations are there in the set {b, c, d, e, f, g} (the original set without ‘a’), we will know how many permutations of the

A powerful tool for solving problems is to transform a given problem into another one that is easier to solve, or that we already know how to solve. Then, from the solution to that new problem, we can give the solution to the former one. This is such a case. We know that we have

“A permutation of a set of distinct objects is an ordered arrangement of these objects.” Theorem: “If n is a positive integer and r is an integer with 1 ≤ r ≤ n, then there are P (n, r ) = n(n − 1)(n − 2) · · · (n − r + 1) r-permutations

This type of exercise is quite easy, and probably you won’t find it commonly on a test, exam, or in real life. However, there are some places you might find it. So, it is worth looking into it. As usual, let’s start with relevant definitions and theorems. “A permutation of a set of distinct objects