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CCM 4300 Lecture 1 Computer Networks: Wireless and Mobile Communication Systems. Permutations and Combinations Dr E. Ever School of Engineering and Information.

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Presentation on theme: "CCM 4300 Lecture 1 Computer Networks: Wireless and Mobile Communication Systems. Permutations and Combinations Dr E. Ever School of Engineering and Information."— Presentation transcript:

1 CCM 4300 Lecture 1 Computer Networks: Wireless and Mobile Communication Systems. Permutations and Combinations Dr E. Ever School of Engineering and Information Sciences

2 Definitions in brief Permutation of a number of objects is the number of different ways they can be ordered. Combinations, one does not consider the order in which objects were chosen or placed. Permutations, position important Combinations, chosen important.

3 A simple example Suppose you have to put some pictures on the wall, and suppose you only have two pictures: A and B. You could hang them in order : Or In this case, for permutations the order of events is important: order 1 is different from order 2. For combinations, it does not matter which picture was hung first (A, B = B, A). A B A B

4 Lets try it with three pictures (1) A B C A C B B A C C B A C B A C A B

5 For the first picture you had a choice of three. For the second picture you had a choice of two. Then you have the last picture. Written algebraically this is 3 × 2 × 1. For large numbers this is factorial notation is used. Lets try it with three pictures (2) What happens if there are 10 pictures and three of them can be hanged? For the first picture you had a choice of ten. For the second picture you had a choice of nine. Then you have eight pictures remaining for the last position. Written algebraically this is 10 × 9 × 8= 720.

6 Formula for the number of possible permutations 10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 3628800. To choose 3 pictures from a set of 10 pictures the first three terms (10 × 9 × 8) are used. The remaining ones (7 × 6 × 5 × 4 × 3 × 2 × 1) are not used. The formula for the number of possible permutations of k objects from a set of n:

7 Combinations (1) The order of the pictures is not important. Consider the hanging of the three pictures A,B,C, we had six permutations. Three pictures were the same in the each permutation They were only one combination. The three pictures made one set, and the same is true for each group of three pictures.

8 What happens if there are 10 pictures and three of them can be hanged? If we take the first three pictures (1, 2 and 3) we could arrange these in six different ways (permutations) 3! Likewise for all the other sets of three pictures; each set of three can be arranged in six different ways. Since each set of three chosen can be arranged in 3! different ways. The number of sets of three (i.e. the number of combinations) is 720 ÷ 3! = 120. Combinations (2)

9 Combinations (3) The formula for the number of possible combinations of r objects from a set of n objects.


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