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© The School For Excellence 2011 Trial Exam Preparation Lectures – Maths Extension 1 – Book 1 Page 20

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

( 1)!n 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? (c) Two particular men are not to sit together?

F

F

F

F

M

M

M

M

© The School For Excellence 2011 Trial Exam Preparation Lectures – Maths Extension 1 – Book 1 1

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


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:

202 190

2!P

In general, the number of ways to select r objects from n where order of selection is not important is:

!

nn r

r

PC

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 105 252C .

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 61 2 3 4 5

The expression nCr can be extended to

!!( )!

nr

nC

r n r

You should prove that expression.


QUESTION 12 In how many ways can a committee of 5 people be selected from 6 men and 8 women if: (a) There are no restrictions? (b) There must be 3 women and 2 men on the committee? (c) A particular man and woman are not allowed to serve on the same committee? (d) There must be at least one woman on the committee? Solution (a) Total number of committees is 14

5 2002C


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?


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?


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?


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


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)?


QUESTION 21

The number of arrangements of 2 2n 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


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 BINOMIAL THEOREM

Consider 2( )x y . It has two terms – therefore we refer to it as a binomial.

The expansion of 2 2 2( ) 2x y x xy y and this has three terms – which is one more than the power.

Similarly another binomial is 3 3 2 2 3( ) 3 3x y x x y xy y 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 2x 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 31C ).

Using this type of approach, we can therefore write out an expansion for a general Binomial expression to the degree n as follows:

1 1 2 20 1 2( ) . . . n n n n n n n n n

na b C a C a b C a b C b

0( )

nn n n r r

rr

a b C a b

We will often have the situation where a = 1 and b = x and the expansion of this is:

2 30 1 2 3(1 ) . . . n n n n n n n

nx C C x C x C x C x

Try this one yourself (and check to see if you are correct):

4(2 3) = When we really want to, we can simplify the long expression into one term with a Ʃ sign in front:

0(1 )

nn n r

rr

x C x

Note the pattern of the coefficients and numbers. Very important and we will come back to this pattern.


Note the following points about the binomial expansion: 1. 0 1n n

nC C

2. n nr n rC C 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 )nx .

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 rx is the 1 thr term in the expansion since the starting value of r is 0.


EXAMPLE 11

Expand 5(2 )x and then write it in sigma notation. Solution (2+x)5 = 5C025x0+ 5C124x1 + 5C223x2 +5C322x3 +5C421x4 +5C520x5 2 3 4 532 80 80 40 10x x x x x Checking the pattern of numbers in the full expansion gives the statement in sigma notation as:

(2 + x)5 = k = 0

5

5Ck (2)5 – k xk

EXAMPLE 12

Expand 4(5 2 )x 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:

(3 – 2x)4 = k = 0

4

4Ck (3)4 – k (-2x)k

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 34 – 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!!!!.


QUESTION 23 Expand:

(a) 53 1x

(b) 4

2 3x

x

(c)

513xy

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