People who buy tickets for a raffle or a lottery discuss their chances of winning. A child in a mother’s womb has equal hances of being a male or female. In all of these situations, you are involved in the chances that the events you discuss will happen. Furthermore, you make estimates, forecasts, or predictions about these events. In this lesson, you will learn about the sample space, events, and getting the probability of an event.

1. P R A Y E R
Dear Lord,
You hold the full of creation in your hands, from the huge
and awe-inspiring universe, to every little grain of sand.
You're the creator of all time, you balance night and day, 𝒊𝒏
𝒇𝒊𝒏𝒊𝒕𝒆 and safe.
We ask for your guidance, with your 𝒂𝒓𝒓𝒐𝒘𝒔 𝒐𝒇 𝒔𝒖𝒄𝒄𝒆𝒔𝒔, so
that we will 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕𝒍𝒚 have a 𝒓𝒂𝒕𝒊𝒐𝒏𝒂𝒍 and 𝒑𝒐𝒔𝒊𝒕𝒊𝒗𝒆
attitude in everything we do.
Give us the 𝒑𝒐𝒘𝒆𝒓 to overcome challenges and 𝒓𝒐𝒐𝒕 out the
bad things in our hearts.
All these we pray in recognition of your power and love,

2. FIVE MINUTES TO WIN IT
Give the end behavior and turning points of the graph.

3. FIVE MINUTES TO WIN IT
Give the end behavior and turning points of the graph.

4. FIVE MINUTES TO WIN IT
Give the end behavior and turning points of the graph.
Behavior:
straight line, passing through the origin,
slanting to the right
Turning Points:
none or zero

5. FIVE MINUTES TO WIN IT
Give the end behavior and turning points of the graph.
Behavior:
rises to the right, rises to the left
Turning Points:
1

6. FIVE MINUTES TO WIN IT
Give the end behavior and turning points of the graph.
Behavior:
rises to the right, falls to the left
Turning Points:
2

7. FIVE MINUTES TO WIN IT
Give the end behavior and turning points of the graph.
Behavior:
both ends up!
Turning Points:
3

8. FIVE MINUTES TO WIN IT
Give the end behavior and turning points of the graph.

9. FIVE MINUTES TO WIN IT
Give the end behavior and turning points of the graph.
Behavior:
straight line, passing through the origin,
slanting to the left
Turning Points:
none or zero

10. FIVE MINUTES TO WIN IT
Give the end behavior and turning points of the graph.
Behavior:
falls to the left, falls to the right
Turning Points:
1

11. FIVE MINUTES TO WIN IT
Give the end behavior and turning points of the graph.
Behavior:
rises to the left, falls to the right
Turning Points:
2

12. FIVE MINUTES TO WIN IT
Give the end behavior and turning points of the graph.
Behavior:
both ends down!
Turning Points:
3

13. PROBLEM SOLVING
INVOLVING
PERMUTATION AND
COMBINATION

14. Objectives:
1. Determine if the given problem is, permutation
or combination.
2. Solve problem involving permutation and
combination.

15. REVIEW
Suppose you are the owner of a sari-sari store and you
want to put 12 pieces of canned goods in a row on the
shelf. If there are 3 identical cans of meat loaf, 4
identical cans of tomato sauce, 2 identical cans of
sardines, and 3 identical cans of corned beef, in how
many different ways can you display these goods?

16. Solution:
Since there are identical elements/objects, we will
use the formula of distinguishable permutation.
P = n!/a!b!c!...
n= 12, a=3, b=4, c=2, d=3
P= 12!/(3!•4!•2!•3!)
P= 277, 200 ways

17. ACTIVITY
1. Find the numbers of different ways of placing 8 marbles in a
row given that 3 are red, 2 are green, 2 yellow, and 1 is black.
2. From the Grade 10 - Soliman, 8 boys and 4 girls decided to
visit their favorite peryahan to experience its number one
attraction, the Merry-Go-Round with 12 horses. How many
ways can students be seated if:
a. no restrictions are imposed
b. all the boys must sit together, and all the girls must
sit together as well

18. Solution for #1:
Keyword: “different ways of placing, repeated element”
in placing, order of colors is important.
Type of Problem: Permutation with Repeated Elements
Given: 𝑛 = 8 𝑛1 = 3 𝑛2 = 2 𝑛3 = 2 𝑛4 = 1
Formula: 𝑃 =
𝑛!
𝑛1!𝑛2!𝑛3!…𝑛𝑘!
Solving: 𝑃 =
8!
3!2!2!1!
=
40,320
24
= 1,680 ways

19. Solution for #2:
Type of Problem: Circular Permutation
a. No restrictions are imposed
Given: 𝑛 = 8
Formula: 𝑃 = 𝑛 − 1 !
Solving: 𝑃 = 12 − 1 ! = 11! = 39,916,880 ways

20. Solution for #2:
Type of Problem: Circular Permutation
b. All the boys must sit together, and all the girls
must sit together as well.
Boys must sit together: 8!
Girls must sit together: 4!
FPC: 8! 4! = 967,880 ways
Treat the 8 boys as 1 group as
well as the 4 girls as 1 group.

21. APPLICATION
1. How many polygons can be possibly formed from 6 distinct
points on a plane, no three of which are collinear?
2. A basket contains 4 star apples, 5 mangoes, and 8 guavas.
How many ways can 2 star apples, 1 mango, and 2 guavas be
chosen?
3. Find the total number of diagonals
that can be drawn in a hexagon.

22. Solution for #1:
The polygon may have 3, 4, 5, or 6 vertices. Thus, the number
N of possible polygons is:
𝑁 = 𝐶 6,3 + 𝐶 6,4 + 𝐶 6,5 + 𝐶 6,6
=
6⋅5⋅4
3!
+
6⋅5⋅4⋅3
4!
+
6⋅5⋅4⋅3⋅2
5!
+
6⋅5⋅4⋅3⋅2⋅1
6!
=
6⋅5⋅4
3⋅1
+
6⋅5⋅4⋅3
4⋅3⋅2⋅1
+
6⋅5⋅4⋅3⋅2
5⋅4⋅3⋅2⋅1
+
6⋅5⋅4⋅3⋅2⋅1
6⋅5⋅4⋅3⋅2⋅1
= 5 ⋅ 4 + 5 ⋅ 3 + 6 + 1
= 20 + 15 + 7
N = 42 possible polygons

23. Solution for #2:
This involves the product of three combinations, one
for each type of item.
4C2 Two of 4 star apples will be chosen
5C1 One of 5 mangoes will be chosen
8C2 Two of 8 guavas will be chosen

24. Solution for #2:
4C2 ∙ 5C1 ∙ 8C2 =
4!
4−2 !2!
⋅
5!
5−1 !1!
⋅
8!
8−2 !2!
=
4!
2!2!
⋅
5!
4!1!
⋅
8!
6!2!
=
4⋅3⋅2!
2⋅2!
⋅
5⋅4!
4!
⋅
8⋅7⋅6!
6!⋅2
= 840

25. Solution for #3:
Each diagonal has two endpoints. Suppose one has endpoints
A and C. Since 𝐴𝐶 and 𝐶𝐴 are the same, order is not important.
The combination of 6 points taken 2 at a time gives the total
number of segments connecting any two points.
6C2=
6!
6−2 !2!
= or 15
However, 15 is not the answer because six of the segments
connecting the points sides of the hexagon. Subtract 6 from 15,
the total number of combinations.
Thus, the total number of diagonals in a hexagon is 15 – 6 or 9.

26. ASSIGNMENT
A circle has eleven randomly-placed points. In how many
ways can you form each polygon listed below?
a. triangle
b. quadrilateral
c. pentagon
d. hexagon
e. octagon
f. decagon

27. Any questions