Math 166: Topics in Contemporary Mathematics II


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1 Math 166: Topics in Contemporary Mathematics II Xin Ma Texas A&M University September 30, 2017 Xin Ma (TAMU) Math 166 September 30, / 11
2 Last Time Factorials For any natural number n, we define n! = n(n 1)(n 2) particularly, 0! = 1 Number of Permutations The number of permutations of r distinct objects taken from a set of size n is given by P(n, r) = n(n 1)(n 2) (n r + 1) = n! (n r)! Particularly, P(n, n) = n(n 1)(n 2) 2 1 = n! Xin Ma (TAMU) Math 166 September 30, / 11
3 Combinations Given the set a, b, c we know that there are P(3, 2) = 6 ways of selecting two of the these at a time when order is important. They are (a, b), (b, a), (a, c), (c, a), (b, c), (c, b). How many ways are there if the order is not important? In this case, (a, b) is the same as (b, a), (a, c) is the same as (c, a) and (b, c) is the same as (c, b). Thus, there are only three ways of selecting two objects when the order doesn t matter and they are (a, b), (a, c) and (b, c). Combinations A combination of the r distinct objects taken form a set having n elements is a selection of r of the objects (without concern for order). Xin Ma (TAMU) Math 166 September 30, / 11
4 Number of Combinations We know the number of permutations of r objects taken form a set of size n is given by P(n, r). Similarly, we denote the number of the combinations of r objects taken form a set of size n by C(n, r). Xin Ma (TAMU) Math 166 September 30, / 11
5 Number of Combinations We know the number of permutations of r objects taken form a set of size n is given by P(n, r). Similarly, we denote the number of the combinations of r objects taken form a set of size n by C(n, r). Number of Combinations The number of combinations of r distinct objects taken from a set of size n is given by C(n, r) = P(n, r) r! = n! (r!)(n r)! Xin Ma (TAMU) Math 166 September 30, / 11
6 Number of Combinations We know the number of permutations of r objects taken form a set of size n is given by P(n, r). Similarly, we denote the number of the combinations of r objects taken form a set of size n by C(n, r). Number of Combinations The number of combinations of r distinct objects taken from a set of size n is given by C(n, r) = P(n, r) r! = n! (r!)(n r)! Example: Find C(7, 3) and C(7, 4). Xin Ma (TAMU) Math 166 September 30, / 11
7 Number of Combinations We know the number of permutations of r objects taken form a set of size n is given by P(n, r). Similarly, we denote the number of the combinations of r objects taken form a set of size n by C(n, r). Number of Combinations The number of combinations of r distinct objects taken from a set of size n is given by C(n, r) = P(n, r) r! = n! (r!)(n r)! Example: Find C(7, 3) and C(7, 4). Note: By the formula above, we have C(n, r) = C(n, n r). Xin Ma (TAMU) Math 166 September 30, / 11
8 Number of Combinations Calculating C(n, r) Using Your Calculator: 1 Input the total number of objects in your set, n 2 Press MATH 3 Scroll to PRB 4 Select ncr (Option 3) 5 Input the number of objects being selected, r 6 Press ENTER Example: Find P(20, 6) and C(20, 6) Xin Ma (TAMU) Math 166 September 30, / 11
9 Number of Combinations Calculating C(n, r) Using Your Calculator: 1 Input the total number of objects in your set, n 2 Press MATH 3 Scroll to PRB 4 Select ncr (Option 3) 5 Input the number of objects being selected, r 6 Press ENTER Example: Find P(20, 6) and C(20, 6) P(20, 6) = and C(20, 6) = Xin Ma (TAMU) Math 166 September 30, / 11
10 a. Papa John s pizza has 18 selections for toppings. How many ways are there to order a 3topping pizza? Xin Ma (TAMU) Math 166 September 30, / 11
11 a. Papa John s pizza has 18 selections for toppings. How many ways are there to order a 3topping pizza? By the definition of combination numbers, there are C(18, 3) = 816 many choices. b. How many different 5card poker hands can be dealt from a standard deck of 52 cards? Xin Ma (TAMU) Math 166 September 30, / 11
12 a. Papa John s pizza has 18 selections for toppings. How many ways are there to order a 3topping pizza? By the definition of combination numbers, there are C(18, 3) = 816 many choices. b. How many different 5card poker hands can be dealt from a standard deck of 52 cards? C(52, 5) = c. How many different flushes (5 cards in a same suit) can be dealt from a standard deck of 52 cards? Xin Ma (TAMU) Math 166 September 30, / 11
13 a. Papa John s pizza has 18 selections for toppings. How many ways are there to order a 3topping pizza? By the definition of combination numbers, there are C(18, 3) = 816 many choices. b. How many different 5card poker hands can be dealt from a standard deck of 52 cards? C(52, 5) = c. How many different flushes (5 cards in a same suit) can be dealt from a standard deck of 52 cards? We view this as a sequence of two tasks. The first one is to choose a suit (spade, diamond, club and heart). Then choose 5 cards from all cards of the suit, which has C(13, 5) many cases. Therefore, we have 4 C(13, 5) = 5148 many cases. Xin Ma (TAMU) Math 166 September 30, / 11
14 a. Find the number of poker hands with three queens and two jacks. Xin Ma (TAMU) Math 166 September 30, / 11
15 a. Find the number of poker hands with three queens and two jacks. We view this as a sequence of two tasks. The first one is to choose 3 queens from 4 queens, which has C(4, 3) = 4 cases. The second one is to choose 2 jacks from 4 jacks, which has C(4, 2) = 6 cases. Therefore, we have 4 6 = 24 cases in total. b. A committee of 15 people need to elect a chair, two associate chairs and three secretaries. How many ways can this be done if no one can hold multiple positions? Xin Ma (TAMU) Math 166 September 30, / 11
16 a. Find the number of poker hands with three queens and two jacks. We view this as a sequence of two tasks. The first one is to choose 3 queens from 4 queens, which has C(4, 3) = 4 cases. The second one is to choose 2 jacks from 4 jacks, which has C(4, 2) = 6 cases. Therefore, we have 4 6 = 24 cases in total. b. A committee of 15 people need to elect a chair, two associate chairs and three secretaries. How many ways can this be done if no one can hold multiple positions? We view this as a sequence of three tasks. The first one is to choose 1 chair from 15 people. There is C(15, 1). Then we choose two associate chairs from rest 14 people: C(14, 2). At last, we choose 3 people from rest 12 people: C(12, 3). Then in total we have C(15, 1)C(14, 2)C(12, 3) = ways to do. Xin Ma (TAMU) Math 166 September 30, / 11
17 a. Find the number of poker hands with three queens and two jacks. We view this as a sequence of two tasks. The first one is to choose 3 queens from 4 queens, which has C(4, 3) = 4 cases. The second one is to choose 2 jacks from 4 jacks, which has C(4, 2) = 6 cases. Therefore, we have 4 6 = 24 cases in total. b. A committee of 15 people need to elect a chair, two associate chairs and three secretaries. How many ways can this be done if no one can hold multiple positions? We view this as a sequence of three tasks. The first one is to choose 1 chair from 15 people. There is C(15, 1). Then we choose two associate chairs from rest 14 people: C(14, 2). At last, we choose 3 people from rest 12 people: C(12, 3). Then in total we have C(15, 1)C(14, 2)C(12, 3) = ways to do. NOTE: it is similar to permutation. Why we cannot use P(15, 5)? Xin Ma (TAMU) Math 166 September 30, / 11
18 A committee of 15 people need to choose a chair, an associate chair and a secretary. How many ways can this be done? Xin Ma (TAMU) Math 166 September 30, / 11
19 A committee of 15 people need to choose a chair, an associate chair and a secretary. How many ways can this be done? P(15, 3) = = Xin Ma (TAMU) Math 166 September 30, / 11
20 A committee of 15 people need to choose a chair, an associate chair and a secretary. How many ways can this be done? P(15, 3) = = By the same way: C(15, 1)C(14, 1)C(13, 1) = = 2730 = P(15, 3). Xin Ma (TAMU) Math 166 September 30, / 11
21 A committee of 15 people need to choose a chair, an associate chair and a secretary. How many ways can this be done? P(15, 3) = = By the same way: C(15, 1)C(14, 1)C(13, 1) = = 2730 = P(15, 3). Why we can use permutation for this example? What is the difference between this one and the question above. Xin Ma (TAMU) Math 166 September 30, / 11
22 A committee of 20 people consists 12 men and 8 women. In how many ways can a subcommittee of 7 be formed if the subcommittee consists of a. any 7 committee members? Xin Ma (TAMU) Math 166 September 30, / 11
23 A committee of 20 people consists 12 men and 8 women. In how many ways can a subcommittee of 7 be formed if the subcommittee consists of a. any 7 committee members? C(20, 7) = b. all men? Xin Ma (TAMU) Math 166 September 30, / 11
24 A committee of 20 people consists 12 men and 8 women. In how many ways can a subcommittee of 7 be formed if the subcommittee consists of a. any 7 committee members? C(20, 7) = b. all men? C(12, 7) = 792. c. only one woman? Xin Ma (TAMU) Math 166 September 30, / 11
25 A committee of 20 people consists 12 men and 8 women. In how many ways can a subcommittee of 7 be formed if the subcommittee consists of a. any 7 committee members? C(20, 7) = b. all men? C(12, 7) = 792. c. only one woman? C(8, 1)C(12, 6) = 7392 Xin Ma (TAMU) Math 166 September 30, / 11
26 d. at least two women? Xin Ma (TAMU) Math 166 September 30, / 11
27 d. at least two women? Think about set theory! What we are doing is counting a number n(a) of a set A. Then union law helps a lot to solve problems of this type and it is helpful to realize this to learn section 2.3 Xin Ma (TAMU) Math 166 September 30, / 11
28 d. at least two women? Think about set theory! What we are doing is counting a number n(a) of a set A. Then union law helps a lot to solve problems of this type and it is helpful to realize this to learn section 2.3 (a) gives the number that we have many ways to choose such a committee without any restriction. (b) gives the number of ways that the committee has ZERO women. (c) gives the number of ways that the committee has exactly ONE women. Thus, by Union Law we have C(20, 7) C(12, 7) C(8, 1)C(12, 6) = many ways to choose a committee with at least two women. Xin Ma (TAMU) Math 166 September 30, / 11
29 Suppose a box contains 9 red balls, 7 blue balls and 8 black balls. How many ways we can pick 7 balls such that there are at least 4 blue balls? Xin Ma (TAMU) Math 166 September 30, / 11
30 Suppose a box contains 9 red balls, 7 blue balls and 8 black balls. How many ways we can pick 7 balls such that there are at least 4 blue balls? We have two ways: (Venn diagram) 1. C(7, 4)C(17, 3) + C(7, 5)C(17, 2) + C(7, 6)C(17, 1) + C(7, 7) Xin Ma (TAMU) Math 166 September 30, / 11
31 Suppose a box contains 9 red balls, 7 blue balls and 8 black balls. How many ways we can pick 7 balls such that there are at least 4 blue balls? We have two ways: (Venn diagram) 1. C(7, 4)C(17, 3) + C(7, 5)C(17, 2) + C(7, 6)C(17, 1) + C(7, 7) 2. C(24, 7) C(17, 7) C(17, 6)C(7, 1) C(17, 5)C(7, 2) C(17, 4)C(7, 3) Xin Ma (TAMU) Math 166 September 30, / 11
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