bhnbhn

2. To count numbers is not obvious and easy.
You don’t think so?
Let’s try then…
1) What is the number of students in the class?
2)What is the number of fingers in your left hand?
3)What is the number of faces of a dice?

4. What about the following questions:
1) How many 5 digit numbers can be formed with odd digits?
2) In how many different ways can 4 pencils be placed in three drawers?
3) How many triangles can be formed by 20 different points?

5. CONCLUSION:
Counting the possibilities of a given situation
may be sometimes easy, but in other cases,
some principle are required

6. Activity 1:
A class is formed of 5 students :
SAMI, MONA, LINA, CARLOS, and HANI
Three students are going to be chosen to represent the school in
three different competitions.
In how many ways can the 3 students be chosen? (If a student can
participate in at most one competition)

7. H
C
L
M
S
H
C
M
S
H
C
S
First
competitor
third
competitor
second
competitor
5 choices
3 choices
(for each preceding choice)
4 choices
(for each preceding choice)
5 × 4 × 3 = 60
then there are 60 possibilities for 3 students to be chosen

8. CONCLUSION:
In general, if in a situation three choices should be made successively, the
first can be made in n different ways, the second in (n – 1) and the third in
(n – 2) ways.
Then the total number of possibilities (choices) is given by :
n × (n-1) × (n – 2)

9. Arrangement:
The number of choices of P elements out of n elements is called arrangement
And it can be calculated using the formula :
Ex:
Calculate

10. Permutation:
The arrangement of n elements out of n elements is called ‘permutation’
It can be calculated by using the arrangement formula just by replacing p by n,
so we can obtain the following expression

11. Understood?????
Then who can give an example of a permutation case?
:D

12. Return to the question of activity 1:
Now in how many ways can the 3 students be chosen If a
student can participate in more than one competition?
In this case, we have 5 choices for the first competitor, and for each
of this choices we have another 5 choices for the second competitor
and 5 choices for the third.
Then the total number of choices is
5 × 5 × 5 = 53 = 125 choices

13. In general , a set of p elements out of n elements , not
necessarily distinct (with repetition) is called p-list
and this number can be calculated by using the
relation np

15. EXERCISES