Hole 5 Hole 8 Distance m,ft 14 Hole Out par 30 Hole 10 11 ...
Hole 1
description
Transcript of Hole 1
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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
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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!!!
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How many different ways can you arrange: 5 bricks in a line, each of a different
colour.
120 ways
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How many arrangements are there of: 5 bricks in a line, where 3 of them are red
and 2 are blue
10 ways
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Classwork
Ex 1A p10
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Binomial Expansion Revision
Consider the expansion of 3qp
qpqpqpqp 3
3223 33 qpqqpp
Why does the expansion have the symmetry in its coefficients?
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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
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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
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)(
!!
! rnrqprnr
n
The general term for the binomial expansion nqp
for r = 0, 1, …, n
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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
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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
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