PROBABILITY PART B.pptx

# If the order of selection matters the selection is

• pauloffei201440
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If the order of selection matters, the selection is called a permutation .

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k -permutations We start with n distinct objects, and let k be some positive integer, with k ≤ n . We wish to count the number of different ways that we can pick k out of these n objects and arrange them in a sequence, i.e., the number of distinct k -object sequences. We can choose any of the n objects to be the first one. Having chosen the first, there are only n − 1 possible choices for the second; given the choice of the first two, there only remain n − 2 available objects for the third stage, etc. When we are ready to select the last (the k th) object, we have already chosen k − 1 objects, which leaves us with n − ( k − 1) choices for the last one. By the Counting Principle
k -permutations the number of possible sequences, called k - permutations , is n ( n − 1) · · · ( n − k + 1) = n ( n − 1) · · · ( n − k + 1)( n − k ) · · · 2 · 1/ ( n − k ) · · · 2 · 1 = n ! / ( n − k )! .= n P r In the special case where k = n , the number of possible sequences, simply called permutations , is n · ( n − 1) · ( n − 2) · · · 2 · 1 = n ! . Note: 0! =1 and 1! = 1

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Example. Let us count the number of words that consist of four distinct letters. This is the problem of counting the number of 4-permutations of the 26 letters in the alphabet. The desired number is n ! /( n − k )! = 26! / 22! = 26 · 25 · 24 · 23 = 358 , 800 .
Example. You have n 1 classical music CDs, n 2 rock music CDs, and n 3country music CDs. In how many different ways can you arrange them so that the CDs of the same type are contiguous? We break down the problem in two stages, where we first select the order of the CD types, and then the order of the CDs of each type. There are 3! ordered sequences of the types of CDs (such as classical/rock/country,rock/country/classical, etc), and there are n 1! (or n 2!, or n 3!) permutations of the classical (or rock, or country, respectively) CDs Thus for each of the 3! CD type sequences, there are n 1! n 2! n 3! arrangements of CDs, and the desired total number is 3! n 1! n 2! n 3! .

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COMBINATIONS Notice that forming a combination is different than forming a k -permutation, because in a combination there is no ordering of the selected elements . The number of combinations of n distinct objects taken r at a time is nCr or = n!/r!(n-r)! Note that: ab is the same combination as ba which is a different permutation. Example 1.28. The number of combinations of two out of the four letters A, B, C, and D is found by letting n = 4 and k = 2. It is 4 C 2

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Example. A class consisting of 4 graduate and 12 undergraduate students is randomly divided into four groups of 4. What is the probability that each group includes a graduate student? We first determine the nature of the sample space. A typical outcome is a particular way of partitioning the 16 students into four groups of 4.
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