6/9/2016MATH 106, Section 51 Section 5 Combinations Questions about homework? Submit homework!

6/9/2016MATH 106, Section 52 What are combinations? Sometimes, we want to take a collection of objects where the order doesn’t matter. Such unordered arrangements are called combinations. 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(5, 2) = 5! ——— = 20 (5 – 2)! P, S S, P P, T T, P P, V V, P P, K K, P S, T T, S S, V V, S S, K K, S T, V V, T T, K K, T V, K K, V Suppose we decide to list all the possibilities! #1

6/9/2016MATH 106, Section 53 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(5, 2) = 5! ——— = 20 (5 – 2)! P, S S, P P, T T, P P, V V, P P, K K, P S, T T, S S, V V, S S, K K, S T, V V, T T, K K, T V, K K, V Suppose we decide to list all the possibilities! Notice that each combination of 2 courses can be obtained by grouping together pairs of permutations of 2 courses. #1

6/9/2016MATH 106, Section 54 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(5, 2) = 5! ——— = 20 (5 – 2)! P, S S, P P, T T, P P, V V, P P, K K, P S, T T, S S, V V, S S, K K, S T, V V, T T, K K, T V, K K, V Suppose we decide to list all the possibilities! 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, c  (2!) = P(5, 2) which implies c = P(5, 2) ——— = 2! 5! ———— = 10 2! (5 – 2)!

6/9/2016MATH 106, Section 55 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 P(30, 4) = 30! ———— = 657,720 (30 – 4)! Let the number of such combinations be c. 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) c = P(30, 4) ——— = 4! 30! ———— = 27,405 4! (30 – 4)! #2

6/9/2016MATH 106, Section 56 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. Let’s generalize this … Whatever handout problems we do not finish here in class we will do in class next time …

6/9/2016MATH 106, Section 57 How many 4-card hands can be drawn from a standard deck of 52 cards? C(52,4) = P(52, 4) ——— = 4! 52! ———— = 270,725 4! (52 – 4)! #4 How many ways can a committee of 5 be chosen from an organization of 40 people? C(40,5) = P(40, 5) ——— = 5! 40! ———— = 658,008 5! (40 – 5)! #3

6/9/2016MATH 106, Section 58 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? C(22,4) = P(22, 4) ——— = 4! 22! ———— = ! (22 – 4)! C(10,2)  C(12,2) = 45  66 = 2970 Choose a committee of 2 girls and 2 boys or … #5

6/9/2016MATH 106, Section 59 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   = 4620

6/9/2016MATH 106, Section 510 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? C(65,8) = P(65, 8) ——— = 8! 65! ———— = 5,047,381,560 8! (65 – 8)! C(15,8) = P(15, 8) ——— = 8! 15! ———— = ! (15 – 8)! C(15,6)  C(50,2) = 5005  1225 = 6,131,125 #6

6/9/2016MATH 106, Section 511 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? C(10,0)  C(55,8) + C(10,1)  C(55,7) + C(10,2)  C(55,6) + C(10,3)  C(55,5) + C(10,4)  C(55,4) C(10,3)  C(5,3)  C(50,3) = 120  10  = 23,520,000 Just set up the calculation.

6/9/2016MATH 106, Section 512 Questions? Be sure to work through Example 4 from the text. (page 41)

6/9/2016MATH 106, Section 513 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, 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. the new employee Jody has been counted in the given information.