Name: Permutations / Combinations March 17, 2014

Transcription

1 1. A pizza establishment offers 12 kinds of meat topping (pepperoni, sausage, etc.) and 5 kinds of vegetable toppings (onions, peppers, etc). How many different two topping pizzas can be made using a combination of 1 meat topping and 1 vegetable topping? 2. Many restaurants offer five-course meals consisting of a soup, salad, appetizer, main course, and dessert. If such a restaurant has three unique soups, two unique salads, five unique appetizers, three unique main course options, and six unique desserts (including Pumpkin Pi), how many unique five-course meals could be created? 3. Christian assembled 5 toy train cars on a track. How many different ways could he have ordered the cars? 4. How many different ways can the letters in the word OLYMPIAD be arranged? 5. In an 8-person race, how many different ways can the first 3 runners arrive at the finish line? 6. In how many ways can the letters in the word BUBBLE be uniquely arranged? 7. The volleyball team has 9 players, but only 6 can be on the court at one time. How many different ways can the team fill the court? 8. On a piano, chords are often played by pressing down three out of twelve different keys. If no key can be struck twice during a given chord, how many different chords are possible? Cascade Ridge PTSA Math Club 1 Show your work on a separate sheet of paper.

2 BONUS PROBLEMS 9. Mr. and Mrs. Smith wanted to have a bit of fun naming their first child, so they decided to choose the letters for their child's name randomly. The only requirements they had for the name were that it is three letters long, that the first and last letters are vowels (a, e, i, o, or u), and that the middle letter is a consonant. How many unique names could that poor child have been given? 10. Sitting in front of you is a locked safe containing $1,000,000. If you know that the five digits in the code to unlock the safe are 2, 3, 3, 8, and 9, how many unique codes are possible? 11. The lock on your diary allows you to set a three-digit combination. To make it harder on the rest of your family though, you decide that each digit must be different. How many possible combinations do you have to choose from? 12. How many ways are there to select a student body president, vice-president, and treasurer from a group of eight candidates? Cascade Ridge PTSA Math Club 2 Show your work on a separate sheet of paper.

3 SOLUTIONS: For each of the 12 possible meat toppings, there are 5 possible vegetable toppings. So, 12 5 = = 540 There are five possiblities for the first car. Once you ve chosen the first car, there are four remaining possibilities for the second car. Then there are three remaining possibilities for the third card. Then two, and then one. So, = 120 This is a permutation, since the order of the cars matters. The total possible permutations when choosing 5 cars out of a set of 5 cars is: npk = 5! = = 120 5! is pronounced 5 factorial 4. 40,320 There are 8 choices for the first letter, then 7 remaining choices for the 2 nd letter, then 6, then 5, etc. So, = 40,320 This is a permutation since the order of the letters matters. The total possible permutations when choosing 8 letters out of a set of 8 letters is: npk = 8! = = 40,320 8! is pronounced 8 factorial The 1 st place winner could be any of the 8 different runners. The 2 nd place winner could be any of the 7 remaining runners. And the 3 rd place winner could be any of the 6 remaining winners. So, Cascade Ridge PTSA Math Club 3 Show your work on a separate sheet of paper.

4 8 7 6 = 336 This is a permutation problem, since the order of the runners matters. We re choosing 3 runners out of total set of 8 runners. So, npk = 8! (8 3)! = 8! 5! = = For the moment, assume that each of the three B s are actually different in some way. We could refer to them as B 1, B 2, and B 3. If this is the case, then there are 6 possibilities for the first letter, 5 for the second, then 4, then 3, etc. So, = 6! = 720 But the reality is that the three B s are actually the same exact letter. The word B 1 UB 2 B 3 LE is really the same as the word B 3 UB 2 B 1 LE. But in our calculation above, we ve actually counted these two variations of BUBBLE as separate words, which is wrong. In fact, in our calculation above, we ve actually counted every word multiple times because of the situation with the three B s. The question is... how many times did we count each unique word? We need to figure out how many different ways the three B s can be ordered. The first B in any given word could be one of three possibilites, the second B could be any one of the two remaining possibilities, and the final B could only be the one remaining choice. So, there are = 6 different orderings for the letter B. And that s exactly how many times we counted each unique word. So, the final correct answer is = 120 This is a problem of combinations, since we don t care about the order of the 6 chosen players. In other words, choosing players A,B,C,D,E,F on the court is considered the same as choosing players F,E,D,C,B,A. But still, we must start with the total permutations (where order matters) and then later undo the effects of the ordering. So, if order mattered, then there are 9 choices for the first player, 8 choices for the second player, then 7, 6, 5, and 4. (Remember we re only choosing 6 players. So, we re choosing 6 out of a set of 9: = 60,480 9! npk = = 60,480 (9 6)! But we ve just counted every combination multiple times. We have counted A,B,C,D,E,F as being a different set than B,A,C,D,E,F, when really they re the same set of 6 players. So we have to undo the effects of this by figuring out how many different possible orderings of six players exist. There are 6 possibilities for the first player, 5 for the second, 4 for the third, etc. So, = 6! = 720 Cascade Ridge PTSA Math Club 4 Show your work on a separate sheet of paper.

5 So, in our original answer (60,480), we actually counted each set of six players 720 times. To get the actual correct answer, 60, = 84 Or, nck = npk k! = 9! (9 6)! 1 6! = The first key in the chord could be any of 12 keys. The second key could be any of the remaining 11. And the third key could be any of the remaining 10. So, the total chords would be: = 1320 But because the order of the three keys doesn t matter when playing a chord, we ve actually counted each chord multiple times. How many times? = 6 times. So, we must divide our answer by 6: = = 220 nck = npk k! = 5! 2! = ! npk = = 720 (10 3)! ! npk = = 336 (8 3)! Cascade Ridge PTSA Math Club 5 Show your work on a separate sheet of paper.