Pascal's Triangle 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 This is the start of Pascal's triangle. It follows a pattern, do you know or can you work out the pattern?

Pascal's Triangle 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 This is the pattern and it can continue forever! We use this to expand brackets- lets see how? + +

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

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

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 1 2 1 1 3 3 1 1 4 6 4 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

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 1 2 1 1 3 3 1 1 4 6 4 1 Power of 0 Power of 1 Power of 2 Power of 3 Power of 4

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

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

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

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

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

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 1 2 1 1 3 3 1 1 4 6 4 1 Power of 0 Power of 1 Power of 2 Power of 3 Power of 4

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 Power of 1 Power of 2 Power of 3 Power of 4 Power of 4 so the coefficients will be: 1 4 6 4 1

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 Power of 1 Power of 2 Power of 3 Power of 4 Power of 4 so the coefficients will be: 1 4 6 4 1 Terms will be: (2x)4, (2x)3(-5), (2x)2(-5)2, (2x)(-5)3 ,(-5)4

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 Power of 1 Power of 2 Power of 3 Power of 4 Power of 4 so the coefficients will be: 1 4 6 4 1 Terms will be: (2x)4, (2x)3(-5), (2x)2(-5)2, (2x)(-5)3 ,(-5)4 16x4, -40x3 , 100x2 , -250x , 625

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 Power of 1 Power of 2 Power of 3 Power of 4 Power of 4 so the coefficients will be: 1 4 6 4 1 Terms will be: (2x)4, (2x)3(-5), (2x)2(-5)2, (2x)(-5)3 ,(-5)4 16x4, -40x3 , 100x2 , -250x , 625 1 4 6 4 1 16x4 - 160x3 + 600x2 - 1000x + 625

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 1 2 1 1 3 3 1 1 4 6 4 1 Power of 0 Power of 1 Power of 2 Power of 3 Power of 4 Power of 4 so the coefficients will be: 1 4 6 4 1 Terms will be: (2x)4, (2x)3(-5), (2x)2(-5)2, (2x)(-5)3 ,(-5)4 16x4, -40x3 , 100x2 , -250x , 625 1 4 6 4 1 16x4 - 160x3 + 600x2 - 1000x + 625 Pg 79 Q 1

Other Questions (2-cx)3 The coefficient of x2 is 294 1 1 1 1 2 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 Power of 0 Power of 1 Power of 2 Power of 3 Power of 4

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 3 3 1 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 Power of 0 Power of 1 Power of 2 Power of 3 Power of 4

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 3 3 1 Terms will be: (2)3, (2)2(-cx), (2)(-cx)2, (cx)3 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 Power of 0 Power of 1 Power of 2 Power of 3 Power of 4

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 3 3 1 Terms will be: (2)3, (2)2(-cx), (2)(-cx)2, (cx)3 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 Power of 0 Power of 1 Power of 2 Power of 3 Power of 4

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 3 3 1 Terms will be: (2)3, (2)2(-cx), (2)(-cx)2, (cx)3 1 3 3 1 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 Power of 0 Power of 1 Power of 2 Power of 3 Power of 4

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: 1 3 3 1 Terms will be: (2)3, (2)2(-cx), (2)(-cx)2, (cx)3 1 3 3 1 6c2x2 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 Power of 0 Power of 1 Power of 2 Power of 3 Power of 4

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

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

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

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?

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

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 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

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

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

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

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 3 x 2 x 1 = 6

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 3 x 2 x 1 = 6 This can be written as 3!  We say ‘three factorial’ NOTICE- ORDER MATTERS  ABC is Different to CBA

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 3 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!

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

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

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

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 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

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

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 =6 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

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:

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

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

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!

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! 3x2

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

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

(n-r)!r! n options r choices 5 options 3 Choices So 5x4x3 = 5x4x3 =10 2 COMBINATION RULE: n options r choices n! (n-r)!r!

(n-r)!r! n options r choices 5 options 3 Choices So 5x4x3 = 5x4x3 =10 2 COMBINATION RULE: n options r choices n! (n-r)!r! 5 options 3 Choices So 5x4x3x2x1 2x1x3x2x1

(n-r)!r! n options r choices 5 options 3 Choices So 5x4x3 = 5x4x3 =10 2 COMBINATION RULE: n options r choices n! (n-r)!r! 5 options 3 Choices So 5x4x3x2x1 2x1x3x2x1

(n-r)!r! n options r choices 5 options 3 Choices So 5x4x3 = 5x4x3 =10 2 COMBINATION RULE: n options r choices n! (n-r)!r! 5 options 3 Choices So 5x4x3x2x1 2x1x3x2x1 3 options 2 Choices So 3x2x1 1x2x1

(n-r)!r! n options r choices 5 options 3 Choices So 5x4x3 = 5x4x3 =10 2 COMBINATION RULE: n options r choices n! (n-r)!r! 5 options 3 Choices So 5x4x3x2x1 2x1x3x2x1 3 options 2 Choices So 3x2x1 1x2x1

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’

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!

How to use this for binomial expansion Eg. (1+x)4 ncr Power of 4 so the coefficients will be: 4c0 14 4c1 13 x1 4c2 12x2 4c3 11x3 4c4 x4

How to use this for binomial expansion Eg. (1+x)4 ncr Power of 4 so the coefficients will be: 4c0 14 4c1 13 x1 4c2 12x2 4c3 11x3 4c4 x4 Notice: The powers of the two numbers add up to = n

How to use this for binomial expansion Eg. (1+x)4 ncr Power of 4 so the coefficients will be: 4c0 14 4c1 13 x1 4c2 12x2 4c3 11x3 4c4 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

How to use this for binomial expansion ncr Power of 4 so the coefficients will be: 4c0 4c1 4c2 4c3 4c4 1 4 6 4 1

How to use this for binomial expansion ncr Power of 4 so the coefficients will be: 4c0 4c1 4c2 4c3 4c4 1 4 6 4 1 Terms will be (2x)4, (2x)3(-5), (2x)2(-5)2, (2x)(-5)3 ,(-5)4 16x4, -40x3 , 100x2 , -250x , 625

How to use this for binomial expansion ncr Power of 4 so the coefficients will be: 4c0 4c1 4c2 4c3 4c4 1 4 6 4 1 Terms will be: (2x)4, (2x)3(-5), (2x)2(-5)2, (2x)(-5)3 ,(-5)4 16x4, -40x3 , 100x2 , -250x , 625 Put them together 1 4 6 4 1 16x4 - 160x3 + 600x2 - 1000x + 625