In a classroom there is a whiteboard and a line of four holes for keeping the marker pens for use on the board. How many different arrangements are there for the pens?

Hole 1 Hole 2 Hole 3 Hole 4

4 choices

3 choices

2 choices

1 choices

4 x 3 x 2 x 1 = 24 different arrangements for the pens

4! = 24

No. arrangements = n! = n(n – 1)(n – 2)…1

3 Coin example: 3 coins are chosen from a bag

C1, C2 and C3

How many different ways are there of choosing them?

C1 C2 C3

C1 and C2 have the same value!!!

How many different ways can you arrange: 5 bricks in a line, each of a different

colour.

120 ways

How many arrangements are there of: 5 bricks in a line, where 3 of them are red

and 2 are blue

10 ways

Classwork

Ex 1A p10

Binomial Expansion Revision

Consider the expansion of 3qp

qpqpqpqp 3

3223 33 qpqqpp

Why does the expansion have the symmetry in its coefficients?

qpqpqpqp 3

If we take one term from each of the brackets and multiply them together, the possible arrangements are:

3223 33 qpqqpp

ppp = p3

ppq = p2q

pqq = pq2

pqp = p2q

qqq = q3

qqp = pq2

qpp = p2q

qpq = pq2

qpqpqpqp 3

We can use factorial notation to find the coefficients of each term:

3223 33 qpqqpp

p3q0 p2q1

1!0!3

!3 3

!1!2

!3

p0q3p1q2

3!2!1

!3 1

!3!0

!3

)3(

!3!

!3 rrqprr

In general, a term for this expansion can be written as

)(

!!

! rnrqprnr

n

The general term for the binomial expansion nqp

for r = 0, 1, …, n

Example 1.

Find the binomial expansion of (p + q)5

Term Coefficient

p5

p4q

p3q2

p2q3

pq4

q5

1!0!5

!5

5!1!4

!5

10!2!3

!5

10!3!2

!5

5!4!1

!5

1!5!0

!5

543223455 510105 qpqqpqpqppqp

Example 1.

Find the term in the expansion of (p + q)12 with p7.

Home workComplete any 6 questions from Ex1B

The required term will be of the form Kp7q5

K = 792!5!7

!12

So the term = 792p7q5

Class workExercise 1B p12 Questions: 1 – 10