Standard DA-5.2 Objective: Apply permutations and combinations to find the number of possibilities of an outcome.

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Presentation transcript:

Standard DA-5.2 Objective: Apply permutations and combinations to find the number of possibilities of an outcome

Objectives: apply fundamental counting principle compute permutations compute combinations distinguish permutations vs combinations

Key Words 1. fundamental principle 2. permutations 3. combinations 4. factorial

Fundamental Counting Principle Fundamental Counting Principle can be used determine the number of possible outcomes when there are two or more characteristics . Fundamental Counting Principle states that if an event has m possible outcomes and another independent event has n possible outcomes, then there are m* n possible outcomes for the two events together.

Fundamental counting principle In a sequence of three events if the first event can happen in ‘m’ ways and the second event in ‘n’ ways and the third event in ‘p’ ways then the sequence can happen in ‘m’ x ‘n’ x ‘p’ ways. This can be extended to any number of events.

Example 1 There are 12 boys and 17 girls in my 2nd block. I need one student from girls and one student from boys to represent my class in my absence. In how many ways I can do this?

Example 2 There are 12 boys and 17 girls in my 2nd block. I need one student from girls or one student from boys to represent my class in my absence. In how many ways I can do this?

Fundamental Counting Principle Lets start with a simple example. A student is to roll a die and flip a coin. How many possible outcomes will there be? 1H 2H 3H 4H 5H 6H 1T 2T 3T 4T 5T 6T 12 outcomes 6*2 = 12 outcomes

Factorial formula For any counting number ‘n’ n factorial = n x (n-1) x (n-2) x (n-3) x ….x 2 x 1 Example:

Permutations Notice, ORDER MATTERS! A Permutation is an arrangement of items in a particular order. Notice, ORDER MATTERS! To find the number of Permutations of n items, we can use the Fundamental Counting Principle or factorial notation.

Permutations The number of ways to arrange the letters ABC: ____ ____ ____ 3 ____ ____ Number of choices for first blank? 3 2 ___ Number of choices for second blank? Number of choices for third blank? 3 2 1 3*2*1 = 6 3! = 3*2*1 = 6 ABC ACB BAC BCA CAB CBA

Permutations To find the number of Permutations of n items chosen r at a time, you can use the formula

Permutations Practice: A combination lock will open when the right choice of three numbers (from 1 to 30, inclusive) is selected. How many different lock combinations are possible assuming no number is repeated?

Permutations Practice: From a club of 24 members, a President, Vice President, Secretary, Treasurer and Historian are to be elected. In how many ways can the offices be filled? Answer Now

Permutations Practice: From a club of 24 members, a President, Vice President, Secretary, Treasurer and Historian are to be elected. In how many ways can the offices be filled?

Combinations ORDER DOES NOT MATTER! A Combination is an arrangement of items in which order does not matter. ORDER DOES NOT MATTER! Since the order does not matter in combinations, there are fewer combinations than permutations.  The combinations are a "subset" of the permutations.

Combinations To find the number of Combinations of n items chosen r at a time, you can use the formula

Combinations To find the number of Combinations of n items chosen r at a time, you can use the formula

Combinations Practice: To play a particular card game, each player is dealt five cards from a standard deck of 52 cards. How many different hands are possible? Answer Now

Combinations To play a particular card game, each player is dealt five cards from a standard deck of 52 cards. How many different hands are possible? Practice:

Combinations Practice: A student must answer 3 out of 5 essay questions on a test. In how many different ways can the student select the questions? Answer Now

Combinations A student must answer 3 out of 5 essay questions on a test. In how many different ways can the student select the questions? Practice:

Combinations Practice: A basketball team consists of two centers, five forwards, and four guards. In how many ways can the coach select a starting line up of one center, two forwards, and two guards? Answer Now

Combinations Practice: A basketball team consists of two centers, five forwards, and four guards. In how many ways can the coach select a starting line up of one center, two forwards, and two guards? Center: Forwards: Guards: Thus, the number of ways to select the starting line up is 2*10*6 = 120.

Solve these problems Find 1. ( 9 P 4) 2. (6 P 3) 3. (5 P 2) 4. (7 P 5)

Solve these problems Find 1. ( 9 C 4) 2. (6 C 3) 3. (5 C 2) 4. (7 C 5)