©Texas A&M University Page 1 Last Name: ___________ First Name: _____________ Section 141 - _____ UIN _____________ Counting Part 2 Part I 1. a. Using one RED, one BLUE, one GREEN, and one ORANGE disc, list all the ways you can choose a set of 2 discs if the order in which you choose them does not matter, i.e., choosing RB is the same as choosing BR. The list of 6 has been started for you. RB, RG, RO, ______, ______, ______ b. There is a total of _______ ways to choose 2 of 4 discs, where order doesn’t matter. When order mattered, there were P(4, 2) = 12 ways to arrange them. Now that order doesn’t matter, there are only 6 ways to choose the set of two. 2. a. If you had 5 discs of different colors (R, B, G, O, P), list all the ways that you could choose a set of 4 in which the order that the discs are chosen doesn’t matter, i.e., RBGO is the same as GRBO. ________________________________________________________________________ b. There is a total of _______ ways to choose 4 of 5 discs, where order doesn’t matter. If order mattered, there would be P(5, 4) = 120 ways to arrange these discs. Now that order doesn’t matter, there are only 5 ways to choose the set of four. A selection of objects, where order does NOT matter, is called a combination. The number of combinations of n objects taken r at a time is !(,)!!nCnrrnr Using this notation, the number of ways to choose 2 of the 4 colored discs in 1. is C(4, 2) and the number of ways to choose 4 of 5 colored discs in 2. is C(5, 4). The formula above may be used, but there is a built-in function on your calculator that will calculate the numerical value of C(n, r). Enter the value of n, press MATH, cursor right to the PRB menu, choose option 3, enter the value of r, and press ENTER. Verify your answers to 1. and 2. on your calculator. 3. Find the number of ways to choose 6 crayons from a box of 24 crayons. C(____, ____) = ________ 4. Find the number of ways to choose 3 pens and 4 pencils from a drawer containing 8 pens and 12 pencils. a. In how many ways can the 3 pens be chosen from the 8 in the drawer? C( ____, ____ ) = __________ b. In how many ways can the 4 pencils be chosen from the 12 in the drawer? C( ____, ____ ) =_________ c. In how many ways can both the pens and pencils be chosen? Hint: Use the Multiplication Principle. __________________ x __________________ = ___________________ # of ways to choose pens # of ways to choose pencilsPermutations and Combinations ©Texas A&M University Page 2 Some examples of other situations involving combinations are - Choosing a 3-member committee from a group of 40 people (order does not matter since the 3 people are being chosen for the same job); - Choosing a committee of 3 males and 1 female from a club with 7 male and 8 female members; - Selecting a sample of 8 light bulbs in which 2 are defective from a box of 30 bulbs that contains 10 defectives. Part II 1. In how many ways can two sets of 5 dominoes be selected from a box of 28 dominoes? C( ____, ____ ) x C( ____, ____ ) = ________________________ # of ways to choose # of ways to choose 5 dominoes from 28 5 dominoes from the remaining for the 1st hand 23 dominoes for the 2nd hand Note: C(28, 10) is incorrect since it would give the number of ways to select 10 dominoes from the set of 28, but does not indicate which 5 dominoes go to which hand. 2. Suppose you buy 9 different canned goods at the grocery store and you want to arrange them in your pantry. The bottom shelf fits 9 cans and the top shelf fits 6 cans. a. In how many ways can all 9 be arranged on the bottom shelf? _______________________ b. In how many ways can 6 of the 9 cans be arranged on the top shelf? ______________________ c. Of the 9 cans, 2 are canned vegetables, 3 are canned fruit, and 4 are soups. If you want to arrange all 9 cans on the bottom shelf by type, in how many ways can this be done? One possible arrangement of the 9 cans is the following: 12 1 2 3 1 2 3 4VV FF F S S S S   To create other arrangements, you can rearrange the order in which the groups appear and you can rearrange the cans within each group. For example, another possible arrangement is 3241 21 132SSSSVVFFF . The total number of arrangements is found as follows: _________ x _________ x _________ x _________ = _________________ # of ways to # of ways to # of ways to # of ways to total # of ways to arrange arrange the 3 arrange the 2 arrange the 3 arrange the 4 cans by type groups vegetable cans fruit cans soup cans 3. At a fruit stand there are 15 watermelons, of which 6 are ripe. A sample of 7 watermelons is selected at random. How many samples would have exactly 5 ripe watermelons? a. If 6 of the watermelons are ripe, then _____ are not ripe. 156___totalripenot ripe b. In the sample of 7, if you want to select exactly 5 ripe watermelons, then you will have exactly ______ unripe watermelons in that sample. c. The 5 ripe watermelons in your sample must be chosen from the 6 ripe watermelons at the fruit stand. Similarly, the 2 unripe watermelons in your sample must be chosen from the 9 unripe watermelons.Permutations and Combinations ©Texas A&M University Page 3 C( _____, _____ ) x C( _____, _____ ) = _____________ # of ways to choose 5 # of ways to choose 2 ripe watermelons from unripe watermelons from 6 ripe available 9 unripe available Note: C(6, 5) without C(9, 2) is incorrect since this only accounts for choosing 5 things in your sample. When dealing with counting problems, you must choose everything that is in your sample. Recall the problem situation: At a fruit stand there are 15 watermelons, of which 6 are ripe. A sample of 7 watermelons is selected at random. d. How many samples of 7 would have exactly 4 or exactly 5 ripe watermelons? The word “or” here breaks up this problem into two separate cases, which can be combined using the union rule. n(exactly 4  exactly 5 ripe) = n(exactly 4 ripe) + n(exactly 5 ripe) – n(exactly 4 ripe  exactly 5 ripe) Note: Since you cannot have exactly 4 AND exactly 5 ripe watermelons in the same sample, then n(exactly 4 ripe  exactly 5 ripe) = 0. Therefore, n(exactly 4 ripe  exactly 5 ripe) = C(