Skip to main content

​
Permutations

Consider the possible arrangements of the letters a, b, and c.
Bandingkan susunan yang memungkinkan dari huruf a, b, John c.
The possible arrangements are: abc, acb, bac, bca, cab, cba.
Susunan yang memungkinkan adalah: abc, acb, bac, bca, cab, cba.
If the order of the arrangement is important then we say that each arrangement is a permutation of the three letters. Thus there are six permutations of the three letters.
Jika urutan susunan penting maka kita katakan bahwa setiap susunan adalah permutasi dari tiga huruf. Dengan demikian ada enam permutasi dari tiga huruf.
An arrangement of n distinct objects in a specific order is called a permutation of the objects.
Susunan n objek yang berbeda dalam urutan tertentu disebut permutasi dari objek.
To determine the number of possibilities mathematically, one can use the multiplication rule to get: 3×2×1=63 \times 2 \times 1 = 6 permutations.

​
Permutation Rule

The arrangement of n objects in a specific order using r objects at a time is called a permutation of n objects taken r objects at a time. It is written as nPr_nP_r and the formula is given by     nPr=n!(n−r)!~~~~~_nP_r = \dfrac{n!}{(n-r)!}.

​
Permutations - Example

How many different ways can a chairperson and an assistant chairperson be selected for a research project if there are seven scientists available?
Solution:Number of ways = ‎7P2=7!(7−2)!=7!5!=7×6=42‎_7P_2 = \dfrac{7!}{(7-2)!} = \dfrac{7!}{5!} = 7 \times 6 = 42.

How many different ways can four books be arranged on a shelf if they can be selected from nine books?
Solution:Number of ways = ‎9P4=9!(9−4)!=9!5!=9×8×7×6=3024‎_9P_4 = \dfrac{9!}{(9-4)!} = \dfrac{9!}{5!} = 9 \times 8 \times 7 \times 6 = 3024.

​
Combinations

Consider the possible arrangements of the letters a, b, and c.
Bandingkan susunan yang memungkinkan dari huruf a, b, dan c.
The possible arrangements are: abc, acb, bac, bca, cab, cba.
Susunan yang memungkinkan adalah: abc, acb, bac, bca, cab, cba.
If the order of the arrangement is not important then we say that each arrangement is the same. We say there is one combination of the three letters.
Jika urutan susunan tidak penting, maka kita katakan bahwa setiap susunan adalah sama. Kita katakan ada satu kombinasi dari tiga huruf.

​
Combinations Rule

The number of combinations of of r objects from n objects is denoted by nCr_nC_r and the formula is given by     nCr=n!(n−r)!r!~~~~~_nC_r = \dfrac{n!}{(n-r)!r!}.

​
Combinations - Example

How many combinations of four objects are there taken two at a time?
Solution:Number of combinations = ‎4C2=4!(4−2)!2!=4!2!2!=4×32×1=6‎_4C_2 = \dfrac{4!}{(4-2)!2!} = \dfrac{4!}{2!2!} = \dfrac{4 \times 3}{2 \times 1} = 6.

In order to survey the opinions of customers at local malls, a researcher decides to select 5 malls from a total of 12 malls in a specific geographic area. How many different ways can the selection be made?
Solution:Number of combinations = ‎_12C5=12!(12−5)!5!=12!7!5!=12×11×10×9×85×4×3×2×1=792‎\_{12}C_5 = \dfrac{12!}{(12-5)!5!} = \dfrac{12!}{7!5!} = \dfrac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} = 792.

In a club there are 7 women and 5 men. A committee of 3 women and 2 men is to be chosen. How many different possibilities are there?
Solution:Number of possibilities: (number of ways of selecting 3 women from 7) ×\times (number of ways of selecting 2 men from 5) = ‎7C3×‎5C2=7!(7−3)!3!×5!(5−2)!2!=7!4!3!×5!3!2!=35×10=350‎_7C_3 \times ‎_5C_2 = \dfrac{7!}{(7-3)!3!} \times \dfrac{5!}{(5-2)!2!} = \dfrac{7!}{4!3!} \times \dfrac{5!}{3!2!} = 35 \times 10 = 350.

A committee of 5 people must be selected from 5 men and 8 women. How many ways can the selection be made if there are at least 3 women on the committee?
Solution:The committee can consist of 3 women and 2 men, or 4 women and 1 man, or 5 women. To find the different possibilities, find each separately and then add them: ‎8C3×‎5C2+‎8C4×‎5C1+‎8C5×‎5C0=56×10+70×5+56×1=560+350+56=966‎_8C_3 \times ‎_5C_2 + ‎_8C_4 \times ‎_5C_1 + ‎_8C_5 \times ‎_5C_0 = 56 \times 10 + 70 \times 5 + 56 \times 1 = 560 + 350 + 56 = 966.