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Warm Up

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- Suppose we wanted to sit six people, three men and three women in a row of six chairs. How many ways can we seat them if we must start with a woman and then alternate women and men?
72

- How many ways can we form a license plate if we want three digits followed by two letters?
676000

- Calculate the number of permutations of n objects taken r at a time.
- Use factorial notation to represent the number of permutations of a set of objects.
- Calculate the number of combinations of n objects taken r at a time.
- Apply the theory of permutations and combinations to solve counting problems.

- Permutation – an ordering of distinct objects in a straight line. If we select r different objects from a set of n objects and arrange them in a straight line, this is called a permutation of n objects taken r at a time. The number of permutations of n objects taken r at a time is denoted by P(n,r).

- How many permutations are there of the letters a, b, c, and d?
- 4 x 3 x 2 x 1 = 24
- P(4, 4) = 24

- How many permutations are there of the letters a, b, c, d, e, f, and g if we take the letters three at a time?
- 7 x 6 x 5 = 210
- P(7, 3) = 210

- If n is a counting number, the symbol n!, called n factorial, stands for the product n•(n – 1) • (n – 2) • (n – 3) • … • 2 • 1. We define 0! = 1.
- P(n, r) = n!
(n – r)!

- If we choose r objects from a set of n objects, we say that we are forming a combination of n objects taken r at a time. We are only concerned with choosing a set, but the order is not important. The notation C(n, r) denotes the number of such combinations.
- C(n, r) = P(n, r) = n!
r! r! • (n – r)!

- In working with permutations and combinations, we are choosing r different objects from a set of n objects. The big difference is whether the order of the objects is important. If it is, then we are dealing with a permutation. If not, then we are working with a combination.
- If a problem involves something other than simply choosing different objects, then perhaps we should use the fundamental counting principle.

- You and fifteen of your friends have formed a company. A committee consisting of a president, vice president, and a three-member executive board need to be formed. How many different ways can this committee be formed?
- Choose the president and vice president.
- The order is important. P(16, 2)

- Select the remaining 3 executive members.
- The order is not important. C(14, 3)

- We can do stage one followed by stage two by using the fundamental counting principle:
P(16, 2) x C(14, 3) = 87,360 ways.

- Choose the president and vice president.

- A class has 10 boys and 12 girls. The teacher wants to choose 4 boys and 3 girls to play 7up. How many ways is this possible?
210 x 220 = 46,200

- A committee has 10 people. A president and vice president needs to be chosen. Then a three person board must be picked. How many ways can this be done?
90 x 56 = 5,040

- Each entry in this triangle after row 1 is the sum of the two numbers immediately above it.
- The nth row of Pascal’s triangle counts the subsets of various sizes of an n-element set.
- List the subsets of {1,2,3,4}: O, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, {1,2,3,4}
- Note the sizes of each are 1 of size zero, 4 of size one, 6 of size two, 4 of size three, and 1 of size four. This pattern occurs as the fourth row in Pascal’s triangle.

- The rth entry of the nth row of Pascal’s triangle is C(n, r).
- The fourth row of Pascal’s triangle is 1 4 6 4 1
- Because the leftmost 1 is the zeroth entry in the fourth row, we can write it as C(4, 0). That is to say, C(4, 0) = 1. The first entry in the fourth row is 4, which means that C(4, 1) = 4.

- Classwork: Page 708 (7 – 15 odd, 19, 21, 37, 41)
- Homework: Page 708 (8 – 16 even, 20, 22, 38, 42)

- Classwork: Page 708 (17, 23, 25, 27, 33, 35, 39, 43, 45, 47, 49, 57, 59, 63)
- Homework: Page 708 (18, 24, 26, 28, 34, 36, 40, 46, 50, 58)