There are 13 members on the board of directors for a hospital. In how many different ways can they elect a chair and a vice chair and a secretary?

Answer 1

Here is a problem in permutations where we use a non-replacement set as a base and draw different combinations of a given number of elements of the base to find a solution. What that means is that as we "use" an element of the set of probable choices, we do not replace it when selection of a second element is made. The first element is "used" only once. An individual is going to hold only one office on the board, as the question asked. We have 13 objects taken 3 at a time in non-replacement manner. As you may have guessed, there is a handy expression for finding the resultant in such a case. It is P(n, r). But we need to know the mathematical notation used in permutations. Let's look at that first. If we take a number (an integer) and multiply it by all the integers up to that number, we are finding what is called a factorial. When written in standard form, the factorial of a number n, is n! (read "enn factorial"), and it is equal to n x (n-1) x (n-2) x (n-3) x (n-4) ... x 4 x 3 x 2 x 1 Nothing trick here. Just take a whole positive number and multiply it by all the integers (counting numbers) up to that number. Piece of cake. The expression we mentioned, P(n, r) = P times n! divided by the quantity (n - r)! P = n! / (n - r)! P is the possible permutations, n is the number of elements in the set from which elements are selected, and r is the number of elements selected. Now we're ready to go to work. P(n, r) = P(13, 3) = 13! / (13 - 3)! = 13! /10! = 13 x 12 x 11 x 10! / 10! = 13 x 12 x 11 = 1716 There are 1716 possible ways to select three officers from among 13 candidates. Note that when we wrote out the solution, we used a short cut for 13! / (13 - 3)! and wrote it 13 x 12 x 11 x 10! / 10! so we could factor out 10! and get to the "easy" part (13 x 12 x 11) to discover our answer. Got a link to the Wikipedia post on permutations. (The link is below.) It's worth that little mouse "click" to surf on over and peek at what is there. It isn't that tough to understand. Since you got this problem, you'll doubtless get more like it - and tougher ones, too boot! Please do your homework. It is a must, and it is really not that tough. Good luck with this.