Statistics and Probability CLAST Review Workshop Statistics and Probability 6. Fundamental Counting Principle (skill I D 3)

Overview The Fundamental Counting Principle . . . . Link Permutations . . . . . . . . . Link Combinations . . . . . . . . Link Formulae . . . . . . . . . . . . Link 12/3/2018 Dave Saha

the Fundamental Counting Principle 1. Count the number of steps in a process. This is the number of individual counts you need to obtain in step 2. 2. For each step count the number of ways that it may be completed. 3. Multiply together all the counts from step 2. The final product is the number of ways of completing the entire process. 12/3/2018 Dave Saha

Fundamental Counting Example 1 Two fair, six-sided dice are to be rolled. How many possible outcomes are there? Rolling the first die is one step; rolling the second die is another step. This is true even if you roll both dice at the same time. Each die has six possible outcomes (1 to 6). So the total number of possible outcomes for the two dice is (6)(6) = 36. 1 2 3 4 5 6 1 . . . . . . 2 . . . . . . 3 . . . . . . 4 . . . . . . 5 . . . . . . 6 . . . . . . 12/3/2018 Dave Saha

Fundamental Counting Example 2 A restaurant offers 5 main entrees, 7 side items, and 6 beverages. How many different meals are possible? Selecting a meal consists of 3 steps, choosing one of each item: entrée, side, and beverage. So the total number of possible meals is (5)(7)(6) = 210. MENU Entrée Side Beverage Steak Potatoes Coffee Chicken Squash Hot Tea Shrimp Beans Iced Tea Salmon Carrots Coca-Cola Pork Peas Sprite Okra Milk Spinach Return to Overview 12/3/2018 Dave Saha

A permutation counts the number of unique arrangements of a group. Permutations A permutation counts the number of unique arrangements of a group. This applies any time a prior choice will affect the number of available choices at a later stage. Example: How many arrangements of the 13 diamonds (from a deck of cards) are possible? (13)(12)(11)(10)(9)(8)(7)(6)(5)(4)(3)(2)(1) = 8,302,694,400 12/3/2018 Dave Saha

Explaining Permutation Number of choices for first card. Number of choices for second card. Number of choices for third card. Number of choices for fourth card. etc. (13)(12)(11)(10)(9)(8)(7)(6)(5)(4)(3)(2)(1) = 8,302,694,400 12/3/2018 Dave Saha

Permutations with Like Items Suppose we have a black bag containing six marbles, all indistinguishable from each other except for color. Three of the marbles are red, and there is one each of blue, green and white. In how many distinct ways may the marbles be drawn, one at a time? If all the marbles were distinguishable, the fundamental counting principle tells us that there are (6)(5)(4)(3)(2)(1) = 720 permutations. 12/3/2018 Dave Saha

Explaining Permutations with Like Items But since we can’t distinguish the red marbles from each other, we have to divide the 720 total permutations by the number of ways the red marbles may be arranged, which is (3)(2)(1) = 6. Thus there are 720 ÷ 6 = 120 unique ways of drawing the marbles. Return to Combinations 12/3/2018 Dave Saha

Permutation Formula The number of marbles, six, is n. The number of selections, also six, is r. P(n, r) = P(6, 6) = = P(6, 6) = (6)(5)(4)(3)(2)(1) = 720 For permutation with like items, just divide by the factorial of the number of like items. (Just click the mouse to advance to the challenge!) 6! (6 – 6)! 0! Link to Formulae Return to Overview 12/3/2018 Dave Saha

CHALLENGE How many different letter arrangements are possible from the letters of “Mississippi”? (Hint, there are 11 letters in the name.) 12/3/2018 Dave Saha

SOLUTION to the CHALLENGE How many different letter arrangements are possible from the letters of “Mississippi”? (Hint, there are 11 letters in the name.) 11! (11)(10)(9)(8)(7)(6)(5)(4)(3)(2)(1) 4!·4!·2! (4)(3)(2)(1)·(4)(3)(2)(1)·(2)(1) = 34,650 = The factors in the denominator account for the multiple i’s (4), s’s (4), and p’s (2). Link to Formulae Return to Overview 12/3/2018 Dave Saha

Combinations What if unique arrangement does not matter for the selection? This happens when we want to form a subset of a group. Suppose a company with 12 programmers wants to establish a team of 3 programmers to do a special project. How many different teams could be formed? It matters only that a programmer is selected for the team, not in which order he/she is selected. This is similar to the previous example with the identical red marbles. 12/3/2018 Dave Saha

Explaining Combinations The number of ways of selecting the team of three programmers is the product of the number of ways of making each selection, (12)(11)(10) = 1320, divided by the possible arrangements of the three: (3)(2)(1) = 6. So the possible number of teams is 1320 ÷ 6 = 220 Return to Overview 12/3/2018 Dave Saha

Formulae } next slide Factorial Mathematicians have an abbreviated form for writing successive multiplication like (11)(10)(9)(8)(7)(6)(5)(4)(3)(2)(1). They write 11! (read “eleven factorial”). SPECIAL: 0! = 1 Permutations Combinations } next slide Return to Permutation Return to Challenge 12/3/2018 Dave Saha

Formulae for Permutations and Combinations Factorial -- previous slide Permutations P(n, r) = Combinations C(n, r) = n! (n – r)! n! r!(n – r)! Return to Permutation (This is the last slide.) Return to Overview 12/3/2018 Dave Saha