## Sunday, January 24, 2016

### Division M2 session on Jan 24, 2016

 Instructor: Mr. Ahmed Hefny Attendance: Ayan, Adan, Omar, Baseer Handout: Combinatorics Problems I Problems Solved in Class: 2,9 and 11 Concepts: Counting, Factorials, Permutations and Combinations Student Difficulties: Although the terms factorials, and combinations seemed to ring a bell. There was difficulty solving more general counting problems. I believe the general material may be new to the students. Homework: Attempt rest of problems in the handout, For questions that ask for probabilities, solve for "In how many ways could X happen" instead of "what is the probability of X happening". Notes: Most students had difficulty with problem 1 in the last geometry homework. Winner: Main ideas: * The number of outcomes of a multistep decision sequence is the product of the number of possibilities for each decision. Example 1: Number of ways to order n elements = n * (n-1) * (n-2) * ... * 1 = n! * If there are different alternatives to decide (part of) the outcome, the number of possibilities of these alternatives should be added. * When confused, draw a tree of decisions and then go bottom-up: each leaf represents a single outcome. The number of outcomes from a on-leaf node is that sum of outcomes from its children. Usually, you won;t need to draw the tree down to the leaves. * One strategy to find the number of possible values of part of the outcome you care about (e.g. 4 out of 10 numbers) is to use the following factorization: number of possible full outcomes = number of possible values of the part we care about * number of ways to decide the values of the part we don't care about. The quantity in red is the one we want to compute, the other two are usually easy to compute. Example 2 - number of ways to select r elements from n elements where order matters (nPr): number of ways to order n elements = nPr * number of ways to order the remaining n-r elements --> n! = nPr * (n-r)! --> nPr = n! / (n-r)! Example 2 - number of ways to select r elements from n elements where order does not matter (nCr):number of ways to select r elements from n elements where order matters = nCr * number of ways to order the selected r elements --> nPr = nCr * r! --> nCr = nPr / r! = n!(n-r)! / r!