. In combinations, order does not matter. 1. Given a standard 52card deck, how many different fivecard hands are possible?


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1 Worksheet ombinations (answers) Mr. hvatal A combination of n objects taken r at a time is denoted by n r. In combinations, order does not matter. Playing cards Examples 1. Given a standard 52card deck, how many different fivecard hands are possible? Answer: hoosing five cards of 52 can be expressed as 52 5, or 2,598, In how many hands are all five cards the same color? Answer: First choose one of the two colors. Then choose five cards of that color. 21i 26 5 = 131, In how many hands are all five cards the same suit? Answer: First choose one of the four suits. Then choose five cards of that suit. 41i 13 5 = 5, Four face cards and one that is not a face card? Answer: There are twelve face cards in every deck, and 40 nonface cards. hoose four of the first group and one of the second. 124i 40 = 19, Three queens and two that are not queens? Answer: There are four queens in every deck, and 48 that are not. hoose three of the first group and two of the second. 43i 48 2 = 4,512
2 6. Full house (three of one kind, two of a different kind)? Answer: We ll need to first choose a single card value from among the 13 possible ( 13 ). Then we can choose 3 out of the four possible of that value, ( 4 ). For the remaining pair, we choose a different card value out of the remaining 12 ( 12 ), then two out of four of this kind ( 4 2 ). Three of a kind i two of a different kind 131i 43i 121i 4 2 = 3, Royal flush (10, J, Q, K and ace of the same suit)? Answer: We ll need to first choose a single suit ( 4 ). Then we must choose 5 specific cards of that suit. Each time we choose a card, we have one less cards remaining in that suit from which to choose. 8. At least one heart? 41i 131i 121i 111i 101i 9 = 4 Answer: There are 13 hearts in every deck, and 39 nonhearts. At least one heart includes five possible outcomes: One heart, two hearts, three hearts, four hearts and five hearts. We ll have to add them to get the total number of possibilities. For each, choose the number of hearts and the number of nonhearts. 9. At least one ace? One heart + two hearts + three hearts + four hearts + five hearts 131i i i i i 390 = 2, 023, 203 Answer: At least one ace includes four possible outcomes: One ace, two aces, three aces, and four aces. We ll have to add them to get the total number of possibilities. For each, choose the number of aces and the number of nonaces. One ace + two aces + three aces + four aces 41i i i i 481 = 886, 656
3 10. At most two spades? Answer: At most two spades includes three possible outcomes: No spades, one spade and four nonspades, and two spades and three nonspades. Again, we ll have to add them to get the total number of possibilities. No spades + one spade + two spades 130i i i 393 = 2,357,862 Note: alculating odds The problems above asked you to calculate the total number of possible hands meeting a given rule, such as at least one ace. To calculate the odds of being dealt that hand (such as 1 in 500 ), you can make a quick additional calculation. total possible outcomes The "odds" = 1 in total successful outcomes To calculate the odds of being dealt a full house (Example 6) would then be: 525 The "odds" of a full house = 1 in 3,744 _ = 1 in One can also express the probability as a percentage. total successful outcomes Probability as a percentage = 100 total possible outcomes i Using the same example, the probability of drawing a full house is: 3,744 Probability as a percentage = 100 i % This will of course be true for calculating odds for any of the following problems, as well.
4 Jelly beans A bag contains 14 red jelly beans, 7 green, 9 yellow, and 6 blue. You will note that these problems require exactly the same approach as problems with playing cards. Examples 11. How many combinations of two green, four yellow, and three red jelly beans are possible out of a selection of nine? What are the odds of drawing this combination? Answer: In turn, calculate the possibility of each successful draw. 2 of 7 green i 4 of 9 yellow i 3 of 14 red i 0 of 6 blue 72i 94i143i 6 0 = 963,144 To calculate the odds: ,143, 280 The odds = 1 in = 963, ,144 1 in Ten jelly beans are drawn. How many combinations are possible in which exactly four are not red? What are the odds of drawing this combination, as a percentage? Answer: In order for exactly four of the ten jelly beans to be not red, six must be red. Four nonred jelly beans must be chosen out of the remaining 22 nonred jelly beans. 6 of 14 red i 4 of 22 nonred 146i 22 4 = 21,966,945 To calculate the probability as a percentage: 21,966,945 Probability = i %
5 Miscellaneous 13. Ten students volunteer to help raise money at the senior class car wash. One student will be assigned to handle the money, three to hold up signs, four to wash and two to towel dry. How many different combinations of assignments can be made? Answer: Although the order of selection does not matter, it may help to think of the process sequentially. First, one of the ten is assigned to handle the money ( 10 ). Out of the remaining nine, three are chosen to hold signs ( 9 ). Four of the remaining six will wash and the rest will towel dry. 101i 93i 64i 2 2 = 12, A committee of three people is to be chosen from seven students: Alan, Bill, harlotte, Denise, Ernie, Frank and Georgia. What is the probability (in percent) that: a) Frank is chosen. Answer: The probably that Frank is chosen, or any other person, is simply 3 of 7, or about 43%. But we can use combinations to find the same value. First, select the number of ways to choose Frank ( 1 ), then choose two out of the remaining six that are not Frank ( 6 2 ). 11i 6 2 = 15 To find the probability, divide by the total number of possible combinations ( 7 ) and multiply by Probability = i % b) Alan and Denise are chosen. Answer: Again, try to find the number of successful outcomes, and divide by the total number of outcomes. First, select the number of ways to choose Alan ( 1 ), then Denise ( 1 ). Then calculate the number of combinations of choosing one of the remaining five people that are not Alan and Denise ( 5 ). 11i 11i 5 = 5
6 Now divide by the total number of possible combinations ( 7 ) and multiply by Probability = i % c) Bill is chosen but not Georgia. Answer: First, select the number of ways to choose Bill ( 1 ), then the number of ways to choose not Georgia for the remaining two spots ( 5 2 ). Remember that Bill is not available as not Georgia. 11i 5 2 = 10 Now divide by the total number of possible combinations ( 7 ) and multiply by Probability = i % d) Ernie or harlotte is chosen. Answer: The two keys to remember are that or means add, and that Ernie and harlotte cannot both be on the committee. So we ll need to calculate the combinations for Ernie, then the combinations for harlotte, and subtract the combinations for Ernie and harlotte on the committee together. Just Ernie + just harlotte (Ernie and harlotte together) i + i i i = Now divide by the total number of possible combinations ( 7 ) and multiply by Probability = i %
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