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Chapter 13: Permutations and Combinations
Permutations and Combinations By Radith Sec 3 Journey
Chapter 13: Permutations and Combinations
The Basic Counting Principle • Counting is easy when you have a list of the items to count. It will be more difficult when there is no list or when the number is too large. In this way, we learn more about more efficient ways of counting. It can be either be shown in tree or box diagram Tree Diagram Box Diagram 5
4
3
2
1
Chapter 13: Permutations and Combinations
Permutations • The number of permutations of n different objects is n! = n(n-1)(n-2)x…….x 3 x 2 x 1 • There are two types of permutations: 1. Permutation of r objects from n different objects 2. Permutation with restriction
Chapter 13: Permutations and Combinations Permutation of r objects from n different objects nP
r=
𝑛! 𝑛−𝑟 !
For example: Find the number of 4-letter permutations that can be formed from the letters in the word JAKARTA Solution: 7P
4=
7! 7−4 !
= 840 ways
Chapter 13: Permutations and Combinations
Permutations with Restrictions Example: 5 boys and 7 girls are to form a line. Find the numbers of permutations in which: a) the first two are girls b) the boys are together Solution: a)
7
6
7 x 6 x 10! = 420 ways b) The boys can be permutated in 5! ways This part and the 7 girls then to form 8 objects which can be permutated in 6! Ways So, 5! X 8! = 4 838 400 ways
Chapter 13: Permutations and Combinations
Combinations • Unlike permutations, a combination is any selection of objects where the order of objects is not concerned 𝑛! nC = r 𝑛−𝑟 !𝑟!
Example: A group of 4 members is to be selected from 7 seniors and 4 juniors. Find the number of ways if: a)
There are no restrictions
b)
The group has exactly 3 seniors
c)
The group has at least 1 junior
Solution: a)
11C 4
b)
7C 3
= 330 ways
x 4C1 = 140 ways
c) Number of groups without juniors = 4C0 x 7C4 = 35 ways So, total number of ways of groups with at least 1 junior = 330 - 35 = 295 ways
Chapter 13: Permutations and Combinations