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Section 5 Combinations

Section 5 Combinations. Questions about homework? Submit homework!. #1. Jane has decided to take two out of five courses: ping pong, swimming, tennis, volleyball, and karate. How many ways can she choose to take one course the first semester and another course the second semester?. P, K

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Section 5 Combinations

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  1. Section 5Combinations • Questions about homework? • Submit homework! MATH 106, Section 5

  2. #1 Jane has decided to take two out of five courses: ping pong, swimming, tennis, volleyball, and karate. How many ways can she choose to take one course the first semester and another course the second semester? P, K K, P P, V V, P S, T T, S P, T T, P P, S S, P 5! ——— = 20 (5 – 2)! P(5, 2)= Suppose we decide to list all the possibilities! T, K K, T T, V V, T V, K K, V S, K K, S S, V V, S What are combinations? • Sometimes, we want to take a collection of objects where the order doesn’t matter. Such unordered arrangements are called combinations. MATH 106, Section 5

  3. #1 Jane has decided to take two out of five courses: ping pong, swimming, tennis, volleyball, and karate. How many ways can she choose to take one course the first semester and another course the second semester? How many ways can she choose to take two courses simultaneously next semester? P, K K, P P, V V, P S, T T, S P, T T, P P, S S, P 5! ——— = 20 (5 – 2)! P(5, 2) = Suppose we decide to list all the possibilities! T, K K, T T, V V, T V, K K, V S, K K, S S, V V, S Notice that each combination of 2 courses can be obtained by grouping together pairs of permutations of 2 courses. MATH 106, Section 5

  4. How many ways can she choose to take one course the first semester and another course the second semester? How many ways can she choose to take two courses simultaneously next semester? P, K K, P P, V V, P S, T T, S P, T T, P P, S S, P 5! ——— = 20 (5 – 2)! P(5, 2) = Suppose we decide to list all the possibilities! T, K K, T T, V V, T V, K K, V S, K K, S S, V V, S In other words, if c is the number of combinations, then the number of permutations is equal to the number of possible combinations (c) times the number of permutations of each combination (2!), that is, 5! ———— = 10 2! (5 – 2)! P(5, 2) ——— = 2! c  (2!) = P(5, 2) which implies c = MATH 106, Section 5

  5. #2 A club consists of 30 students. How many ways can we choose students to serve as president, vice president, treasurer, and secretary? How many ways can we choose 4 students to serve on a committee? The number of possibilities we have when how the four students are ordered matters is equal to 30! ———— = 657,720 (30 – 4)! P(30, 4)= and is also equal to the number of possible combinations (c) times the number of permutations of each combination (4!), that is, c  (4!) = P(30, 4) Let the number of such combinations be c. 30! ———— = 27,405 4! (30 – 4)! P(30, 4) ——— = 4! c = MATH 106, Section 5

  6. Let’s generalize this … • P(n, k) is the number of permutations of n things taken k at a time. • C(n, k) is the number of combinations of n things taken k at a time. Whatever handout problems we do not finish here in class we will do in class next time … MATH 106, Section 5

  7. #3 How many ways can a committee of 5 be chosen from an organization of 40 people? 40! ———— = 658,008 5! (40 – 5)! P(40, 5) ——— = 5! C(40,5) = #4 How many 4-card hands can be drawn from a standard deck of 52 cards? 52! ———— = 270,725 4! (52 – 4)! P(52, 4) ——— = 4! C(52,4) = MATH 106, Section 5

  8. #5 A class has 10 girls and 12 boys. How many committees of 4 can be formed? How many committees with 2 girls and 2 boys can be formed? How many committees with at least 2 girls can be formed? 22! ———— = 7315 4! (22 – 4)! P(22, 4) ——— = 4! C(22,4) = 45  66= 2970 C(10,2)  C(12,2)= Choose a committee of 2 girls and 2 boys or … MATH 106, Section 5

  9. How many committees with at least 2 girls can be formed? Choose a committee of 2 girls and 2 boys or Choose a committee of 3 girls and 1 boy or Choose a committee of 4 girls. C(10,2)  C(12,2) + C(10,3)  C(12,1) + C(10,4) = 45  66 + 120  12 + 210 = 4620 MATH 106, Section 5

  10. #6 A summer school has 10 faculty, 5 staff members, and 50 students. How many committees of 8 members can be formed? How many committees of 8 members with no students can be formed? How many committees of 8 members with exactly 2 students can be formed? 65! ———— = 5,047,381,560 8! (65 – 8)! P(65, 8) ——— = 8! C(65,8) = 15! ———— = 6435 8! (15 – 8)! P(15, 8) ——— = 8! C(15,8) = C(15,6)  C(50,2) = 5005  1225 = 6,131,125 MATH 106, Section 5

  11. How many committees of 8 members with at most 4 faculty can be formed? How many committees of 9 members with exactly 3 from each category can be formed? Just set up the calculation. C(10,1)  C(55,7) + C(10,0)  C(55,8) + C(10,2)  C(55,6) + C(10,4)  C(55,4) C(10,3)  C(55,5) + C(10,3)  C(5,3)  C(50,3)= 120  10  19600= 23,520,000 MATH 106, Section 5

  12. Questions? • Be sure to work through Example 4 from the text. (page 41) MATH 106, Section 5

  13. Homework Hints: In Section 5 Homework Problem #10, In Section 5 Homework Problem #10(d), In Section 5 Homework Problem #10(e), In Section 5 Homework Problem #10(f), In Section 5 Homework Problem #11, the new employee Jody has been counted in the given information. notice that “at most 3” means exactly the same as “3 or less”. notice that there is one less committee member to choose. notice that there is one less employee among the personnel to choose committee members from. first write an appropriate recipe to choose the poker hand. MATH 106, Section 5

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