Page 5 - Pra U STPM 2022 Penggal 3 - Maths (T)

Mathematics Semester 3 STPM Chapter 2 Probability

Example 2

A password consists of three symbols. If the first symbol is a character, the second and the third symbols
are digits, how many three-symbol passwords could be formed?
Solution: Let A be the set of characters, B and C be the sets of digits. Then A × B × C is
the set of all passwords fulfilled the requirement. There are 26 elements in the
set of characters, 10 elements in the set of digits. We thus have
n(A) × n(B) × n(C)
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= 26 × 10 × 10
= 2 600
Thus, a total of 2 600 three-symbol passwords could be formed. 2

Example 3

Find the number of different 5 digit numbers. How many of these numbers are even?
Solution: The first digit could be any numbers from 1 to 9. Each of the next four digits
could be any digits. By the multiplication principle, there are
4
9 × 10 = 90 000 such numbers.
Now the final digit must be one of the numbers 0, 2, 4, 6 or 8, i.e. 5 ways. Again
by the multiplication principle, there are
3
9 × 10 × 5 = 45 000 such numbers.
Note: If no letters can be duplicated in a label, then the first letter of a label can be selected from all 26
characters. The second letter, however, must be selected from 25 characters because one letter has
been selected for the first position and that letter cannot be used for the second position. Similarly
the third letter is now selected from the remaining 24 characters and the fourth from 23 characters.

Permutations

Assume that you try to arrange three persons A, B and C sitting in a row for dinner, how many ways can
you line up the three persons? As the number is small, it is not difficult to make such arrangement. We
could write down all the possibilities: ABC, ACB, BCA, BAC, CAB and CBA. But what if there were eight
persons? Now, the arrangement is getting more complicated. We must come up a new counting method
to handle this problem. The way to do this is to work in steps and then use the multiplication principle.
Back to the case of three persons, at the initial stage we have three ways to fill the first seat, then two persons
remain to fill the second seat and finally just one person left for the last seat. So, the number of ways to
arrange the persons in order is 3 × 2 × 1 = 6.
This multiplication principle works with any finite number of persons. For eight persons, the number of
distinct arrangements is 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40 320.
The above discussion introduces another basic counting principle.
A permutation of a set of elements is a way of arranging all or part of the elements in a definite order.

For n elements of a set to be orderly arranged, we can do this in n steps by line up one element at a time.
There are n choices for the first place. Once the first place is filled, any one of the n – 1 elements can be
filled in the second place, and so on. For each step, there is one less than the total elements in the previous
step. By the multiplication principle, the number of permutations is
.
.
n(n – 1)(n – 2) … 3 2 1

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02 STPM Math(T) T3.indd 73 28/10/2021 10:21 AM