Download arrangements in a circle

ARRANGEMENTS IN A CIRCLE
When objects are arranged in a circle, the total number of arrangements is reduced. The
arrangement of (say) four people in a line is easy and no problem (if they listen of course!!).
With a circle, the arrangement of A, B, C, D is exactly the same as the arrangement B, C, D,
A – all we (as the observer or the waiter) have to do is so move one position around and we
have exactly the same arrangement as the first.
A circle does not have a beginning so it is necessary to place one person into their seat and
then arrange the other people as though they were being ordered in a line. So for 4 people,
sit on down and arrange the others (in 3! ways).
In general, the number of ways of arranging n objects in a circle (without restriction) is
( n  1)!
EXAMPLE 8
In how many ways can 4 men and 4 women be seated around a circular table if:
(a)
There are no restrictions?
(b)
The men and women are seated alternately?
M
F
F
M
M
F
F
M
(c)
Two particular men are not to sit together?
© The School For Excellence 2011
Trial Exam Preparation Lectures – Maths Extension 1 – Book 1
Page 20
ARRANGEMENTS WHERE THERE ARE REPETITIONS
The total number of possible arrangements is also reduced when there is more than one of a
particular object being considered. The common example is a word which has repeated
letters. Here any particular arrangement looks to be the same if (say) two letter Es are
interchanged. So what we have to do is to remove the effect of the number of ways the
repetition can be made.
EXAMPLE 9
In how many ways can the letters of the word MATHEMATICS be arranged in a row with no
restrictions?
Solution
There are 11 letters in the word MATHEMATICS, 2 each of M, A and T and one each of the
remainder.
The number of ways to arrange 11 letters is 11! = 39 916 800.
But now, by exchanging the two Ms, we get MATHEMATICS – looks familiar.
As there are 2! ways of arranging the two Ms and every arrangement of the 11 letters has
that duplication, we must divide the total number of arrangements by 2!.
The same logic must apply to the two Ts and to the two As.
 The number ways the letters can be arranged is
11!
 4 989 600 .
2!2!2!
In summary:
If n objects of which p are alike of one type and q are alike of another and so on, are to be
arranged in a row, then the number of arrangements is given by:
n!
p !q !......
© The School For Excellence 2011
Trial Exam Preparation Lectures – Maths Extension 1 – Book 1
Page 21
COMBINATIONS
A combination is the selection of objects from a large group but where order of selection (i.e.
arrangement) is not important. In such a case, we might want to select two students from a
class to represent the class on the SRC. What is important is that a particular two students
are selected – NOT in which order they are selected. So we want Lily and Tom, for example,
and it does not matter which is chosen first.
We think of this as:
(a)
How can we select two students from a class of 20 – 20P2.
(b)
How many ways could those two students be arranged? 2!
Hence the number of ways to select the two students in any order is:
20
P2
 190
2!
In general, the number of ways to select r objects from n where order of selection is not
important is:
n
Cr 
n
Pr
r!
We use the symbol C to represent combinations. The number before the C and above is the
total pool from which we make a selection. The number after and below the C is the number
we a selecting without order being important.
EXAMPLE 10
The number of ways to select 5 people from a group of 10 is
C5  252 .
10
Use the C button on your calculator to obtain the answer. Remember however what the C
stands for. Here it means
10  9  8  7  6
1 2  3  4  5
The expression nCr can be extended to
n
Cr 
n!
r !( n  r )!
You should prove that expression.
© The School For Excellence 2011
Trial Exam Preparation Lectures – Maths Extension 1 – Book 1
Page 22
QUESTION 12
In how many ways can a committee of 5 people be selected from 6 men and 8 women if:
(a)
(b)
(c)
(d)
There are no restrictions?
There must be 3 women and 2 men on the committee?
A particular man and woman are not allowed to serve on the same committee?
There must be at least one woman on the committee?
Solution
(a)
Total number of committees is
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C5  2002
Trial Exam Preparation Lectures – Maths Extension 1 – Book 1
Page 23
QUESTION 13
A team of eleven players is to be selected from a soccer team consisting of 18 players,
where 3 are goalkeepers, 8 are forwards and 7 are defenders. In how many ways can this
be done if:
(a)
There are no restrictions?
(b)
The team must have at least 2 forwards?
(c)
The team is to be made up of a goalkeeper, 4 defenders, 3 midfielders who can be
chosen from any of the forwards and defenders and 3 forward players?
© The School For Excellence 2011
Trial Exam Preparation Lectures – Maths Extension 1 – Book 1
Page 24
QUESTION 14
Three Mathematics books, six French books and four Music books are to be arranged on a
shelf. All the books are different. In how many ways can the books be arranged if:
(a)
There are no restrictions?
(b)
The books from each subject are to be together?
QUESTION 15
(a)
Five people are to travel together to a concert. The car they are to travel in seats 2
people in the front and 3 in the back. If there are 2 drivers, find the number of ways the
5 people can be seated.
(b)
In how many ways can this be done if one of the passengers (not a driver), must have
a window seat?
© The School For Excellence 2011
Trial Exam Preparation Lectures – Maths Extension 1 – Book 1
Page 25
QUESTION 16
How many numbers can be formed with the digits 0, 1, 5, 6, 7, 9 if there are no repetitions
(numbers cannot start with 0) and:
(a)
All digits must be used?
(b)
All digits are used and the number is odd?
(c)
The number must be bigger than 6000?
© The School For Excellence 2011
Trial Exam Preparation Lectures – Maths Extension 1 – Book 1
Page 26
QUESTION 17
Five boys are to sit around a circular table. In how many ways can this be done if:
(a)
There are no restrictions?
(b)
Two of the boys are not allowed to sit together?
QUESTION 18
A registration number plate is to consist of 2 letters followed by 3 digits. Registration plates
starting with the letter Z or the combination CD, are not allowed. How many number plates
are possible?
Solution
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Trial Exam Preparation Lectures – Maths Extension 1 – Book 1
Page 27
QUESTION 19
Six boys and three girls are to be seated in a room where the chairs are set up in two rows
of 6. In how many ways can this be done if:
(a)
The first row must be fully occupied?
(b)
The three girls must sit together?
QUESTION 20
(a)
How many ‘words’ can be made from the letters of the word PERMUTATIONS if all
letters are to be used?
(b)
In how many words will the vowels be together in their alphabetical order (a, e, i, o, u)?
© The School For Excellence 2011
Trial Exam Preparation Lectures – Maths Extension 1 – Book 1
Page 28
QUESTION 21
The number of arrangements of  2n  2  objects taken n at a time to the number of
arrangements of 2n objects taken n at a time is in the ratio 3:1. Find the value of n .
Solution
© The School For Excellence 2011
Trial Exam Preparation Lectures – Maths Extension 1 – Book 1
Page 29
QUESTION 22
A group of people are to play a card game where 5 cards are dealt to each player. Find the
number of 5-card hands that can be dealt which consist of:
(a)
Red cards only.
(b)
No court cards (i.e. king, queen, jack).
(c)
Just one suit.
(d)
At least 2 diamonds.
(e)
The four aces.
© The School For Excellence 2011
Trial Exam Preparation Lectures – Maths Extension 1 – Book 1
Page 30
THE BINOMIAL THEOREM
Consider ( x  y ) 2 . It has two terms – therefore we refer to it as a binomial.
The expansion of ( x  y ) 2  x 2  2 xy  y 2 and this has three terms – which is one more than
the power.
Similarly another binomial is ( x  y )3  x 3  3 x 2 y  3 xy 2  y 3 but its expansion has four terms
(again one more than the power).
The first and last terms are straightforward. For the second term – that in x 2 y – we have two
x terms and one y term. So we have to arrange these – how many ways to arrange three
terms? It is 3 (or 3C1 ).
Using this type of approach, we can therefore write out an expansion for a general Binomial
expression to the degree n as follows:
( a  b ) n  nC0 a n  nC1a n 1b1  nC 2 a n  2 b 2  . . .  nC n b n
n
( a  b ) n   n Cr a n  r b r
r 0
We will often have the situation where a = 1 and b = x and the expansion of this is:
(1  x ) n  nC0  nC1 x  nC 2 x 2  nC3 x 3  . . .  nC n x n
Try this one yourself (and check to see if you are correct):
(2  3) 4 =
When we really want to, we can simplify the long expression into one term with a Ʃ sign in
front:
n
(1  x) n   nCr x r
r 0
Note the pattern of the coefficients and numbers. Very important and we will come back to
this pattern.
© The School For Excellence 2011
Trial Exam Preparation Lectures – Maths Extension 1 – Book 1
Page 31
Note the following points about the binomial expansion:
1.
n
C 0  nC n  1
2.
n
C r  nC n  r i.e. the coefficients are symmetrical.
3.
As the powers of a decrease, the powers of b increase but the sum
of the two powers in any one term is constant and equal to n .
4.
There are n + 1 terms in the expansion of the binomial to the power n - (1  x) n .
5.
As the powers of a decrease, the powers of b increase but the sum of the two powers
in any one term is constant and equal to n .
6.
The term in x r is the  r  1
© The School For Excellence 2011
th
term in the expansion since the starting value of r is 0.
Trial Exam Preparation Lectures – Maths Extension 1 – Book 1
Page 32
EXAMPLE 11
Expand (2  x)5 and then write it in sigma notation.
Solution
(2+x)5 = 5C025x0+ 5C124x1 + 5C223x2 +5C322x3 +5C421x4 +5C520x5
 32  80 x  80 x 2  40 x 3  10 x 4  x 5
Checking the pattern of numbers in the full expansion gives the statement in sigma notation
as:

5
5
(2 + x) =
5
Ck (2)
5–k k
x
k=0
EXAMPLE 12
Expand (5  2 x) 4 and then write it in sigma notation.
Solution
(3-2x)4 = 4C034(-2x)0+ 4C133(-2x)1 + 4C232(-2x)2 +4C331(-2x)3 +4C430(-2x)4
= 81 – 216x + 216x2 – 96x3 + 16x4
In sigma notation:

4
4
(3 – 2x) =
4
Ck (3)
4–k
(-2x)
k
k=0
The expression following the sigma sign can be examined independently because it can be
regarded as being the general term in the expansion.
In the last example, the general term (which can be written as T k+1) is
Tk + 1 = 4Ck 3
4–k
(-2x)
k
If we wish to write down the term which has an x2 component, then k = 2
and it will be the (k+1) = 3rd term.
Note: When writing down the general term, brackets have been placed around the (-2x).
This ensures the negative sign, the 2 and the x are all raised to the power k. Too often,
students take a short cut, forget the brackets and then forget to raise both the negative and
the 2 to the power. What a shame – more marks lost through a lack of precaution. We are
sure no one at TSFX would do that from now on!!!!. 
© The School For Excellence 2011
Trial Exam Preparation Lectures – Maths Extension 1 – Book 1
Page 33
QUESTION 23
Expand:
(a)
 3x  1
(b)
 2 3
x  
x

(c)

1
 3x  
y

5
4
5
© The School For Excellence 2011
Trial Exam Preparation Lectures – Maths Extension 1 – Book 1
Page 34