1. Pascal's Triangle 1
1 1
This is the start of Pascal's triangle. It follows a pattern, do you know or can you work out the pattern?

2. Pascal's Triangle 1
1 1
This is the pattern and it can continue forever!
We use this to expand brackets- lets see how? + +

3. Expand these:
1. (a+b)0
2. (a+b)1
3. (a+b)2
4. (a+b)3
5. (a+b)4

4. Expand these: 1. (a+b)0 2. (a+b)1 3. (a+b)2 (a+b)3 (a+b)4 =1 =1a+1b
=1a2+2ab+1b2
=1a3+3a2b+3ab2+1b3
=1a4+4a3b+6a2b2+4ab3+1b4

5. Expand these: 1 1a+1b 1a2+2ab+1b2 1a3+3a2b+3ab2+1b3
1 1 1
1a+1b
1a2+2ab+1b2
1a3+3a2b+3ab2+1b3
1a4+4a3b+6a2b2+4ab3+1b4

6. Expand these: 1 1. (a+b)0 1 1 2. (a+b)1 1 2 1 3. (a+b)2 1 3 3 1 (a+b)3
Coefficients 1
1 1
Power of 0
1. (a+b)0
2. (a+b)1
3. (a+b)2
(a+b)3
(a+b)4
Power of 1
Power of 2
Power of 3
Power of 4

7. How is this useful? (x+2y)3 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 Power of 0
1 1
Power of 0
Power of 1
Power of 2
Power of 3
Power of 4

8. How is this useful? (x+2y)3 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1
1 1
Power of 0
Power of 1
Power of 2
Power of 3
Power of 4
Power of 3 so the coefficients will be:

9. How is this useful? (x+2y)3 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1
1 1
Power of 0
Power of 1
Power of 2
Power of 3
Power of 4
Power of 3 so the coefficients will be:
Terms will be:
(x)3, (x)2(2y), (x)(2y)2, (2y)3

10. How is this useful? (x+2y)3 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1
1 1
Power of 0
Power of 1
Power of 2
Power of 3
Power of 4
Power of 3 so the coefficients will be:
Terms will be:
(x)3, (x)2(2y), (x)(2y)2, (2y)3
x3 , 2x2y , 4xy , 8y3

11. How is this useful? (x+2y)3 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1
1 1
Power of 0
Power of 1
Power of 2
Power of 3
Power of 4
Power of 3 so the coefficients will be:
Terms will be:
(x)3, (x)2(2y), (x)(2y)2, (2y)3
x3 , 2x2y , 4xy , 8y3

12. How is this useful? (x+2y)3 x3 +6x2y + 12xy2 +8y3 1 1 1 1 2 1 1 3 3 1
1 1
Power of 0
Power of 1
Power of 2
Power of 3
Power of 4
Power of 3 so the coefficients will be:
Terms will be:
(x)3, (x)2(2y), (x)(2y)2, (2y)3
x3 , 2x2y , 4xy , 8y3
x x2y xy2 +8y3

13. How is this useful? (2x-5)4 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 Power of 0
1 1
Power of 0
Power of 1
Power of 2
Power of 3
Power of 4

14. How is this useful?
(2x-5)4 1
1 1
Power of 0
Power of 1
Power of 2
Power of 3
Power of 4
Power of 4 so the coefficients will be:

15. How is this useful?
(2x-5)4 1
1 1
Power of 0
Power of 1
Power of 2
Power of 3
Power of 4
Power of 4 so the coefficients will be:
Terms will be:
(2x)4, (2x)3(-5), (2x)2(-5)2, (2x)(-5)3 ,(-5)4

16. How is this useful?
(2x-5)4 1
1 1
Power of 0
Power of 1
Power of 2
Power of 3
Power of 4
Power of 4 so the coefficients will be:
Terms will be:
(2x)4, (2x)3(-5), (2x)2(-5)2, (2x)(-5)3 ,(-5)4
16x4, -40x3 , 100x , x , 625

17. How is this useful?
(2x-5)4 1
1 1
Power of 0
Power of 1
Power of 2
Power of 3
Power of 4
Power of 4 so the coefficients will be:
Terms will be:
(2x)4, (2x)3(-5), (2x)2(-5)2, (2x)(-5)3 ,(-5)4
16x4, -40x3 , 100x , x , 625
16x x x x

18. How is this useful? Pg 79 Q 1 (2x-5)4 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1
1 1
Power of 0
Power of 1
Power of 2
Power of 3
Power of 4
Power of 4 so the coefficients will be:
Terms will be:
(2x)4, (2x)3(-5), (2x)2(-5)2, (2x)(-5)3 ,(-5)4
16x4, -40x3 , 100x , x , 625
16x x x x
Pg 79 Q 1

19. Other Questions (2-cx)3 The coefficient of x2 is 294 1 1 1 1 2 1
1 1
Power of 0
Power of 1
Power of 2
Power of 3
Power of 4

20. Other Questions (2-cx)3 The coefficient of x2 is 294 1 1 1 1 2 1
Power of 3 so the coefficients will be: 1
1 1
Power of 0
Power of 1
Power of 2
Power of 3
Power of 4

21. Other Questions (2-cx)3 The coefficient of x2 is 294 1 1 1 1 2 1
Power of 3 so the coefficients will be:
Terms will be:
(2)3, (2)2(-cx), (2)(-cx)2, (cx)3 1
1 1
Power of 0
Power of 1
Power of 2
Power of 3
Power of 4

22. Other Questions (2-cx)3 The coefficient of x2 is 294 1 1 1 1 2 1
Power of 3 so the coefficients will be:
Terms will be:
(2)3, (2)2(-cx), (2)(-cx)2, (cx)3 1
1 1
Power of 0
Power of 1
Power of 2
Power of 3
Power of 4

23. Other Questions (2-cx)3 The coefficient of x2 is 294 1 1 1 1 2 1
Power of 3 so the coefficients will be:
Terms will be:
(2)3, (2)2(-cx), (2)(-cx)2, (cx)3 1
1 1
Power of 0
Power of 1
Power of 2
Power of 3
Power of 4

24. Other Questions 6c2x2 (2-cx)3 The coefficient of x2 is 294 1 1 1 1 2 1
Power of 3 so the coefficients will be:
Terms will be:
(2)3, (2)2(-cx), (2)(-cx)2, (cx)3
6c2x2 1
1 1
Power of 0
Power of 1
Power of 2
Power of 3
Power of 4

25. Other Questions 6c2x2 6c2=294 (2-cx)3 The coefficient of x2 is 294 1
Power of 3 so the coefficients will be:
Terms will be:
(2)3, (2)2(-cx), (2)(-cx)2, (cx)3
6c2x2
6c2=294 1
1 1
Power of 0
Power of 1
Power of 2
Power of 3
Power of 4

26. Other Questions 6c2x2 6c2=294 (2-cx)3 The coefficient of x2 is 294
Power of 3 so the coefficients will be:
Terms will be:
(2)3, (2)2(-cx), (2)(-cx)2, (cx)3
6c2x2
6c2=294 1
1 1
Power of 0
Power of 1
Power of 2
Power of 3
Power of 4
c2=49
c=+7 or -7

27. Other Questions 6c2x2 6c2=294 (2-cx)3 The coefficient of x2 is 294
Power of 3 so the coefficients will be:
Terms will be:
(2)3, (2)2(-cx), (2)(-cx)2, (cx)3
6c2x2
6c2=294 1
1 1
Power of 0
Power of 1
Power of 2
Power of 3
Power of 4
c2=49
c=+7 or -7
Pg 79 Q 2

28. 1)Three people were in a race – ABC How may different options of
finishing positions are there?
2)Two out of those 3 people were
going to be picked- the order they are
picked doesn’t matter.
How many outcomes are there for this?

29. 1)Three people were in a race – ABC How may different options of
finishing positions are there?
You may have worked the following options out:
ABC BAC CAB ACB BCA CBA

30. 1)Three people were in a race – ABC How may different options of
finishing positions are there?
You may have worked the following options out:
ABC BAC CAB ACB BCA CBA
So 6 is the answer. However there is a more mathematical method which becomes more helpful with more options
eg. in a race of 20 people

31. 1st 2nd 3rd 1)Three people were in a race – ABC
How may different options of
finishing positions are there?
You may have worked the following options out:
ABC BAC CAB ACB BCA CBA
So 6 is the answer. However there is a more mathematical method which becomes more helpful with more options
eg. in a race of 20 people
1st
2nd
3rd

32. 1st 2nd 3rd 1)Three people were in a race – ABC
How may different options of
finishing positions are there?
You may have worked the following options out:
ABC BAC CAB ACB BCA CBA
So 6 is the answer. However there is a more mathematical method which becomes more helpful with more options
eg. in a race of 20 people
1st
2nd
3rd
3 options for people who could come first

33. 1st 2nd 3rd 1)Three people were in a race – ABC
How may different options of
finishing positions are there?
You may have worked the following options out:
ABC BAC CAB ACB BCA CBA
So 6 is the answer. However there is a more mathematical method which becomes more helpful with more options
eg. in a race of 20 people
1st
2nd
3rd
So only 1 person left for 3rd
3 options for people who could come first
Once 1st is decided only 2 people left for 2nd

34. 1st 2nd 3rd 1)Three people were in a race – ABC
How may different options of
finishing positions are there?
You may have worked the following options out:
ABC BAC CAB ACB BCA CBA
So 6 is the answer. However there is a more mathematical method which becomes more helpful with more options
eg. in a race of 20 people
1st
2nd
3rd
3 options for people who could come first
Once 1st is decided only 2 people left for 2nd

35. 1st 2nd 3rd So 3 x 2 x 1 = 6 3 options of people who could come first
So only 1 person left for 3rd
Once 1st is decided only 2 people left for 2nd
So x 2 x 1 = 6

36. 1st
2nd
3rd
3 options of people who could come first
So only 1 person left for 3rd
Once 1st is decided only 2 people left for 2nd
So x 2 x 1 = 6
This can be written as 3!  We say ‘three factorial’
NOTICE- ORDER MATTERS  ABC is Different to CBA

37. 1st
2nd
3rd
3 options of people who could come first
So only 1 person left for 3rd
Once 1st is decided only 2 people left for 2nd
So x 2 x 1 = 6
This can be written as 3!  We say ‘three factorial’
NOTICE- ORDER MATTERS  ABC is Different to CBA
So in a race of 20 people the number of different outcomes of finishing positions would have 20 options for 1st place, 19 left for second, 18 left for third etc so would be TWENTY FACTORIAL- 20!

38. FACTORIALS
Work out these : 5! 7! 8! 4!
d) 10! 6!

39. FACTORIALS Answers to the : 5! 5x4x3x2x1 =120 7! 7x6x5x4x3x2x1= 5040
d) 10! 10x9x8x7x6x5x4x3x2x1 = 5040
6! 6x5x4x3x2x1

40. 2)Two out of those 3 people were
going to be picked- the order they are
picked doesn’t matter.
How many outcomes are there for this?
You may have worked the following options out:
AB AC BC
Same as BA Same as CA Same as CB

41. 2)Two out of those 3 people were
going to be picked- the order they are
picked doesn’t matter.
How many outcomes are there for this?
You may have worked the following options out:
AB AC BC
Same as BA Same as CA Same as CB
So 3 is the answer. This is a tiny bit harder to work out mathematically:

42. 1st 2nd 2)Two out of those 3 people were
going to be picked- the order they are
picked doesn’t matter.
How many outcomes are there for this?
You may have worked the following options out:
AB AC BC
Same as BA Same as CA Same as CB
So 3 is the answer. This is a tiny bit harder to work out mathematically:
1st
2nd
There are 3 choices for the first pick

43. 1st 2nd 2)Two out of those 3 people were
going to be picked- the order they are
picked doesn’t matter.
How many outcomes are there for this?
You may have worked the following options out:
AB AC BC
Same as BA Same as CA Same as CB
So 3 is the answer. This is a tiny bit harder to work out mathematically:
1st
2nd
There are 3 choices for the first pick
So that leaves 2 choices for the second pick

44. 1st 2nd 2)Two out of those 3 people were
going to be picked- the order they are
picked doesn’t matter.
How many outcomes are there for this?
You may have worked the following options out:
AB AC BC
Same as BA Same as CA Same as CB
So 3 is the answer. This is a tiny bit harder to work out mathematically:
1st
2nd
There are 3 choices for the first pick
So that leaves 2 choices for the second pick
So 3 x 2 = but that gives the options if AB is the same as BA,
AC as CA and BC as CB
But the question states the order doesn’t matter- so 3x2 =3 2

45. How many outcomes are there for this?
3 out of those 5 people were going to be picked- the order they are picked doesn’t matter.
How many outcomes are there for this?
Outcomes:
ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE
So 10 is the answer, but mathematically:

46. How many outcomes are there for this?
3 out of those 5 people were going to be picked- the order they are picked doesn’t matter.
How many outcomes are there for this?
Outcomes:
ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE
So 10 is the answer, but mathematically:
Take ABC as an example if the order mattered these are all the options: ABC ACB
BAC BCA
CAB CBA

47. 3 out of those 5 people were going to be picked- the order they are picked doesn’t matter.
How many outcomes are there for this?
Outcomes:
ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE
So 10 is the answer, but mathematically:
Take ABC as an example if the order mattered these are all the options: ABC ACB
BAC BCA
CAB CBA
1st
2nd
3nd
There are 3 choices
for the third pick
There are 5 choices
for the first pick
So that leaves 4 choices
for the second pick

48. 3 out of those 5 people were going to be picked- the order they are picked doesn’t matter.
How many outcomes are there for this?
Outcomes:
ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE
So 10 is the answer, but mathematically:
Take ABC as an example if the order mattered these are all the options: ABC ACB
BAC BCA
CAB CBA
1st
2nd
3nd
There are 3 choices
for the third pick
There are 5 choices
for the first pick
So that leaves 4 choices
for the second pick
This is like the first question - 3 choices put in order which is 3!

49. 3 out of those 5 people were going to be picked- the order they are picked doesn’t matter.
How many outcomes are there for this?
Outcomes:
ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE
So 10 is the answer, but mathematically:
Take ABC as an example if the order mattered these are all the options: ABC ACB
BAC BCA
CAB CBA
1st
2nd
3nd
There are 3 choices
for the third pick
There are 5 choices
for the first pick
So that leaves 4 choices
for the second pick
This is like the first question - 3 choices put in order which is 3!
So 5 x 4 x 3 =60
But the question states the order doesn’t matter- so 5x4x3 = 5x4x3 =10
3! x2

50. Can you see any rule? 5 options 3 Choices So 5x4x3 = 5x4x3 =10 3! 3x2

51. 5 options 3 Choices
So 5x4x3 = 5x4x3 =10
3! x2
3 options 2 Choices
So 3x2 =3 2

52. (n-r)!r! n options r choices 5 options 3 Choices So 5x4x3 = 5x4x3 =102
COMBINATION RULE:
n options r choices n!
(n-r)!r!

53. (n-r)!r! n options r choices 5 options 3 Choices So 5x4x3 = 5x4x3 =102
COMBINATION RULE:
n options r choices n!
(n-r)!r!
5 options 3 Choices
So 5x4x3x2x1
2x1x3x2x1

54. (n-r)!r! n options r choices 5 options 3 Choices So 5x4x3 = 5x4x3 =102
COMBINATION RULE:
n options r choices n!
(n-r)!r!
5 options 3 Choices
So 5x4x3x2x1
2x1x3x2x1

55. (n-r)!r! n options r choices 5 options 3 Choices So 5x4x3 = 5x4x3 =102
COMBINATION RULE:
n options r choices n!
(n-r)!r!
5 options 3 Choices
So 5x4x3x2x1
2x1x3x2x1
3 options 2 Choices
So 3x2x1
1x2x1

56. (n-r)!r! n options r choices 5 options 3 Choices So 5x4x3 = 5x4x3 =102
COMBINATION RULE:
n options r choices n!
(n-r)!r!
5 options 3 Choices
So 5x4x3x2x1
2x1x3x2x1
3 options 2 Choices
So 3x2x1
1x2x1

57. Combinations can use 2 notations- you need to recognise both!
ncr
n – is the total number of options
r - is the number of choices you will make
Says ‘ n choice r’

58. Combinations can use 2 notations- you need to recognise both!
ncr n r
n – is the total number of options
r - is the number of choices you will make
Says ‘ n choice r’
COMBINATION RULE:
n options r choices n!
(n-r)!r!

59. How to use this for binomial expansion
Eg. (1+x)4
ncr
Power of 4 so the coefficients will be:
4c c1 13 x c2 12x c3 11x c4 x4

60. How to use this for binomial expansion
Eg. (1+x)4
ncr
Power of 4 so the coefficients will be:
4c c1 13 x c2 12x c3 11x c4 x4
Notice:
The powers of the two numbers add up to = n

61. How to use this for binomial expansion
Eg. (1+x)4
ncr
Power of 4 so the coefficients will be:
4c c1 13 x c2 12x c3 11x c4 x4
Notice:
The powers of the two numbers add up to = n
The value of r is the same as the power of the second half of the term

62. How to use this for binomial expansion
ncr
Power of 4 so the coefficients will be:
4c c c c c4

63. How to use this for binomial expansion
ncr
Power of 4 so the coefficients will be:
4c c c c c4
Terms will be
(2x)4, (2x)3(-5), (2x)2(-5)2, (2x)(-5)3 ,(-5)4
16x4, -40x3 , 100x , x , 625

64. How to use this for binomial expansion
ncr
Power of 4 so the coefficients will be:
4c c c c c4
Terms will be:
(2x)4, (2x)3(-5), (2x)2(-5)2, (2x)(-5)3 ,(-5)4
16x4, -40x3 , 100x , x , 625
Put them together
16x x x x