1'1The Fundamental Principle of Multiplication

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Permutations and Combinations

• 1.1 The Fundamental Principle of Multiplication

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The Fundamental Principle of Multiplication

• If there are
• n1 ways of doing one operation,
• n2 ways of doing a second operation, n3 ways of
  doing a third operation , and so forth,

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• then the sequence of k operations can be
  performed in n1 n2 n3.. nk ways.
• N n1 n2 n3.. nk

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Example 1

• A used car wholesaler has agents who classify
  cars by size (full, medium, and compact) and age
  (0 - 2 years, 2- 4 years, 4 - 6 years, and over 6
  years).
• Determine the number of possible automobile
  classifications.

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Solution
The tree diagram enumerates all possible
classifications, the total number of which is
3x4 12.
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Example 2

• Mr. Chan has 2 pairs of trousers, 3 shirts and 2
  ties.
• He chooses a pair of trousers, a shirt and a tie
  to wear everyday.
• Find the maximum number of days he does not need
  to repeat his clothing.

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Solution

• The maximum number of days he does not need to
  repeat his clothing is 232 12

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Class Practice 1.1

• (1)
• If there are 3 routes from town A to town B and 4
  routes from town B to town C,
  how many different
  routes can be taken in getting from A to C via B?

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(2) The chief surgeon for an upcoming transplant
operation is preparing to select the supporting
surgical team. Needed will be one resident
surgeon, one anesthesiologist, one surgical
nurse, one assisting nurse, and one orderly.
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• Given the date of the surgery, the chief surgeon
  can select from 5 resident surgeons, 3
  anesthesiologists, 6 surgical nurses, 10
  assisting nurses and 4 orderlies.
• How many possible supporting surgical teams can
  be selected?

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(3) A cancer research project classifies persons
in four categories male or female heavy
smoker, moderate smoker, or nonsmoker regular
exercise program or no regular program
overweight or not overweight. Enumerate all
possible classifications of persons.
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1.2 Factorials

• The product of the first n consecutive integers
  is denoted by n! and is read as factorial n.
•  That is n! 1234. (n-1) n
• For example,
• 4!1x2x3x424,
•   7!12345675040.
•  Note 0! defined to be 1.

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• The product of any number of consecutive integers
  can be expressed as a quotient of two factorials,
  for example,
• 6789 9!/5! 9! / (9 4)!
• 1112131415 15! / 10!
• 15! / (15 5)!
• In particular,
• n(n 1)(n 2)...(n r 1)
• n! / (n r)!

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• Class Practice 1.2
• 1. Evaluate 8! (a) 5! (b) 7!
• Express in factorial notation (a) 1234,
• (b) 89l011.

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1.3 Permutations

• (A) Permutations
• A permutation is an arrangement of objects.
• abc and bca are two different permutations.
• 1. Permutations with repetition

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• The number of permutations of r objects, taken
  from n unlike objects,
• can be found by considering the number of ways of
  filling r blank spaces in order with the n given
  objects.
• If repetition is allowed, each blank space can be
  filled by the objects in n different ways.

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• Therefore, the number of permutations of r
  objects, taken from n unlike objects,
• each of which may be repeated any number of times
  
• n n n .... n(r factors) nr

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2. Permutations without repetition

• If repetition is not allowed, the number of ways
  of filling each blank space is one less than the
  preceding one.

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• Therefore, the number of permutations of r
  objects, taken from n unlike objects, each of
  which can only be used once in each permutation
• n(n 1)(n2) .... (nr 1)
• Various notations are used to represent the
  number of permutations of a set of n elements
  taken r at a time

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• some of them are

Since
We have
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Example 3

• How many 4-digit numbers can be made from the
  figures 1, 2, 3, 4, 5, 6, 7 when
• (a) repetitions are allowed
• (b) repetition is not allowed?
• Solution
• (a) Number of 4-digit numbers
• 74 2401.
• (b) Number of 4 digit numbers
• 7 6 5 4 840.

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Example 4

• In how many ways can 10 men be arranged
• (a) in a row,
• (b) in a circle? 
• Solution 
• (a) Number of ways is
• 3628800 

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• Suppose we arrange the 4 letters A, B, C and D in
  a circular arrangement as shown.
• Note that the arrangements ABCD, BCDA, CDAB and
  DABC are not distinguishable.

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• For each circular arrangement there are 4
  distinguishable arrangements on a line.
• If there are P circular arrangements, these yield
  4P arrangements on a line, which we know is 4!.

Hence
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Solution (b)

• The number of distinct circular arrangements of n
  objects is (n 1)!
• Hence 10 men can be arranged in a circle in 9!
  362 880 ways.

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Class Practice 1.3

• In how many ways can 10 people be seated in 10
  chairs
• In how many ways can 20 people be seated in 30
  chairs

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• How many 3-digit numbers can be made from the
  figures 1, 2, 3, 4, 5,6, 7, 8, 9 when
• (a) repetitions are allowed,
• (b) repetition is not allowed?

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(B) Conditional Permutations

• When arranging elements in order , certain
  restrictions may apply.
• In such cases the restriction should be dealt
  with first..

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Example 5How many even numerals between 200 and
400 can be formed by using 1, 2, 3, 4, 5 as
digits(a) if any digit may be repeated(b) if
no digit may be repeated?
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• Solution (a)
• Number of ways of choosing the hundreds digit
  2.
• Number of ways of choosing the tens digit 5.
• Number of ways of choosing the unit digit 2. 
• Number of even numerals between 200 and 400 is
• 2 5 2 20.

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• Solution (b)
• If the hundreds digit is 2,
• then the number of ways of choosing an even
  unit digit 1,
• and the number of ways of choosing a tens digit
  3.
• the number of numerals formed
• 113 3.

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• If the hundreds digit is 3, then the number
  of ways of choosing an even. unit digit 2, and
  the number of ways of choosing a tens digit 3.
• number of numerals formed
• 123 6.
• the number of even numerals between 200 and 400
  3 6 9

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Example 6 In how many ways can 7 different books
be arranged on a shelf(a) if two particular
books are together
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• Solution (a)
• If two particular books are together, they can be
  considered as one book for arranging.
• The number of arrangement of 6 books
• 6! 720.
• The two particular books can be arranged in 2
  ways among themselves.
• The number of arrangement of 7 books with two
  particular books together
• 6! x 2 1440.

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(b) if two particular books are separated?

• Solution (b)
• Total number of arrangement of 7 books 7!
  5040.
• the number of arrangement of 7 books with 2
  particular books separated 5040 -1440 3600.

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(C) Permutation with Indistinguishable Elements

• In some sets of elements there may be certain
  members that are indistinguishable from each
  other.
• The example below illustrates how to find the
  number of permutations in this kind of situation.

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Example 7In how many ways can the letters of the
word ISOS CELES be arranged to form a new
word ?

• Solution
• If each of the 9 letters of ISOSCELES were
  different, there would be P 9! different
  possible words.

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• However, the 3 Ss are indistinguishable from
  each other and can be permuted in 3! different
  ways.
• As a result, each of the 9! arrangements of the
  letters of ISOSCELES that would otherwise spell
  a new word will be repeated 3! times.

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• To avoid counting repetitions resulting from the
  3 Ss, we must divide 9! by 3!.
• Similarly, we must divide by 2! to avoid counting
  repetitions resulting from the 2
  indistinguishable Es.
• Hence the total number of words that can be
  formed is
• 9! 3! 2! 30240

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• If a set of n elements has k1 indistinguishable
  elements of one kind, k2 of another kind,
• and so on for r kinds of elements, then the
  number of permutations of the set of n elements is

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Example 8

• The streets of a town form a rectangular grid, 5
  blocks long and 4 blocks wide, as shown in the
  diagram.
• A man walks along the street from P to Q
  always walking either due East or due North.
• (a) Find the total number of possible paths.

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• (b) Find the number of these paths which pass
  through the point R.

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• Solution (a)
• Each path from P to Q involves a walk of 5 blocks
  due East and 4 blocks due North.
• Hence each path corresponds to an arrangement of
  the nine letters EEEEENNNN, taken together.
• The number of paths is equal to the number of
  permutations of the nine letters,
• 9! 5! 4! 126 paths

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• Similarly, the number of paths from P to R
• 5! 4! 1! 5
• and the number of paths from R to Q
• 4! 1! 3! 4
• The total number of paths from P to Q via R
• 5 4 20

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Class Practice 1.4

• (1) Five boys and two girls are to be seated
  in a row. In how many ways can this be done if
• (a) a girl must sit at either end of the row,
• (b) the two girls must not sit next to each
  other.

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• (2)
• How many different numbers can be indicated by
  rearranging the 9 digits of 988877666?

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• (3)
• In how many ways can 3 different Mathematics, 2
  different Physics and 4 different Chemistry books
  be arranged on a shelf if
• the Mathematics books are next to each other,
• the books are kept in subject groupings?

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• (4)
• In how many ways can the letters of the word
  AGREEMENT be arranged to from a new word?

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Exercise 1(a) p9

• 1.(a) A woman has 5 skirts and 6 blouses.
  Assuming that each blouse can be worn with each
  skirt, how many different skirt-blouse outfits
  does the woman have?
• (b) A full dinner at a restaurant consists of
  one of 5 appetizers, one of 6 main courses, and
  one of 8 deserts. How many different complete
  dinners are there?

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Exercise 1(a) p9

• (2)
• In how many ways can 4 people be accommodated in
  5 rooms if they
• (a) are put in separate rooms,(b) dont mind
  sharing?

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Exercise 1(a) p9

• (3)
• In how many ways can 4 different prizes be
  awarded in a class of 15 boys if
• (a) no boy may win more than 1 prize,(b) any
  boy may win all the prizes?

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Exercise 1(a) p9

• (4)
• A man has time to visit one friend on each
  evening of a given week.
• There are 12 friends whom he like to visit.
• In how many ways can he plan his week if
• (a) he can visit a friend more than once,(b)
  he will not visit a friend more than once?

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Exercise 1(a) p9

• (5)
• Using all the digits 1,2,3,4,5,6 how many
  arrangements can be made
• (a) beginning with an even digit,(b) beginning
  and ending with an even digit?

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Exercise 1(a) p9

• (6)
• If women must go first, in how many ways can 5
  women and 7 men enter a lifeboat?

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Exercise 1(a) p9

• (7)
• Five boys and two girls sit in a row. In how many
  ways is this possible if the girls
• (a) must not sit together, (b) must sit at the
  ends?

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Exercise 1(a) p9

• (8)
• Four visitors A, B, C and D arrive in a town
  which has 6 hotels.
• In how many ways can they disperse themselves
  among the 6 hotels
• (a) if four hotels are used to accommodate
  them,(b) if three hotels are used to accommodate
  them in such a waythat A and B stay at the same
  hotel?

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Exercise 1(a) p10

• (9)
• How many numbers between 4000 and 7000 can be
  formed by using 3, 4, 5, 6 as digits, so that any
  digit may be repeated ?

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Exercise 1(a) p10

• (10)
• In how many ways can 4 Mathematics books and 6
  Physics books be arranged on a shelf
• (a) with no restrictions,
• (b) if all the Mathematics books are to the
  left of the Physics books.

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Exercise 1(a) p10

• (11)
• In how many ways can 5 people be seated for a
  photograph
• if there are 2 seats in the front row and 3
  seats in the back row?

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Exercise 1(a) p10

• (12)
• In how many ways can a president, vice-president,
  secretary, and treasurer be selected from a class
  of 30 students, if no person may hold two
  offices?

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Exercise 1(a) p10

• (13)
• If the first 2 symbols in a license tag are
  letters of the alphabet and the next 4 symbols
  are digits of our numeral system, how many
  different tags can be made
• (a) if no symbol may be repeated,
• (b) if any symbol may be repeated?

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Exercise 1(a) p10

• (14)
• A particular new car model is available with 5
  choices of colour, 3 choices of transmission, 4
  types of interior and 3 types of engine. How many
  different cars of this model are possible ?

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Exercise 1(a) p10

• (15)
• Employee ID numbers at a certain factory consist
  of one capital letter followed by a 3-digit
  number containing no repeat digits.
• (For example, A025 is an ID number.)
• How many such ID numbers can be formed ?
• How many could be formed if repeat digits were
  allowed?

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Exercise 1(a) p10

• (16)
• In the figure shown, the vertices represent
  cities and the edge routes between cities. In how
  many different ways can a man starting at A visit
  each other city once and return to A?

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Exercise 1(a) p11

• (17)
• There are 5 boys and 5 girls at a dance.
• In how many different ways can they pair off
  for dancing,
• if everyone is dancing with a member of the
  opposite sex?

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Exercise 1(a) p11

• (18)
• How many different numbers can be indicated by
  rearranging
• (a) the 8 digits of 122 33344?
• (b) the 11 digits of 99988887765?

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Exercise 1(a) p11

• (19)
• A signaler has six flags, of which one is blue,
  two are white and three are red. He sends
  messages by hoisting flags on a flagpole, the
  message being conveyed by the order in which the
  colours are arranged. Find how many different
  message he can send
• (a) by using exactly six flags,
• (b) by using exactly five flags.

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1.4 Combinations

• When a selection of objects is made with no
  regard being paid to order, it is referred to as
  a combination.
• Thus, ABC, ACB, BAG, BCA, CAB, CBA are different
  permutation, but they are the same combination of
  letters.

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• Suppose we wish to appoint a committee of 3 from
  a class of 30 students.
• We know that P330 is the number of different
  ordered sets of 3 students each that may be
  selected from among 30 students.
• However, the ordering of the students on the
  committee has no significance,

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• so our problem is to determine the number of
  three-element unordered subsets that can be
  constructed from a set of 30 elements.
• Any three-element set may be ordered in 3!
  different ways, so P330 is 3! times too large.
• Hence, if we divide P330 by 3!,the result will be
  the number of unordered subsets of 30 elements
  taken 3 at a time.

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• This number of unordered subsets is also called
  the number of combinations of 30 elements taken 3
  at a time, denoted by C330 and

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• In general, each unordered r-element subset of a
  given n-element set (r? n) is called a
  combination.
• The number of combinations of n elements taken r
  at a time is denoted by Cnr or nCr or C(n, r) .
• A general equation relating combinations to
  permutations is

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• Note
• (1) Cnn Cn0 1
• (2) Cn1 n 
• (3) Cnn Cnn-r
• Class Practice 1.5
• 1. Evaluate
• (a) C52 (b) C55 (c) C51
• 2. Evaluate
• (a) C85 (b) C74C75
• 3. Prove that Cnr Cnn-r

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Example 9How many different 5-card hands can be
dealt from a deck of 52 playing cards?

• Solution
• Since we are not concerned with the order in
  which each card is dealt, our problem concerns
  the number of combinations of 52 elements taken 5
  at a time.
• The number of different hands is
• C525 2118760.

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Example 106 points are given and no three of
them are collinear.(a) How many triangles can be
formed by using 3 of the given points as
vertices?

• Solution
• (a) Number of triangles
• number of ways
• of selecting 3 points out of 6
• C63 20.

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(b) How many pairs of triangles can be formed by
using the 6 points as vertices?

• Solution
• Let the points be A, B, C, D, E, F.
• If A, B, C are selected to form a triangles, then
  D, E, F must form the other triangle.
• Similarly, if D, E, F are selected to form a
  triangle, then A, B, C must form the other
  triangle.

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• Therefore, the selections A, B, C and D, E, F
  give the same pair of triangles and the same
  applies to the other selections.
• Thus the number of ways of forming a pair of
  triangles
• C63 2 10

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Example 11From among 25 boys who play
basketball, in how many different ways can a team
of 5 players be selected if one of the players is
to be designated as captain?

• Solution
• A captain may be chosen from any of the 25
  players.
• The remaining 4 players can be chosen in C254
  different ways.
• By the fundamental counting principle, the total
  number of different teams that can be formed is
• 25 C244265650.

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• Alternative Method
• Without designating a captain the number of ways
  would be C255 .
• However, 5 different captains may be chosen in
  each different set of players.
• The total number of different teams is 5
  C255 265650

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Class Practice 1.6

• 1. Write down all 2-element
• subsets that can be constructed
• from the set a,b,c,d,e.
• 2. A secondary school principal
• plans to appoint a committee of 4
• from the 48 teachers. How many
• different committees are possible?
• 3. How many different 13-card hands
• can be dealt from a deck of 52
• playing cards?

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(B) Conditional Combinations

• If a selection is to be restricted in some way,
  this restriction must be dealt with first.
• The following examples illustrate such
  conditional combination problems.

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Example 13A committee of 3 men and 4 women is to
be selected from 6 men and 9 women. If there is
a married couple among the 15 persons, in how
many ways can the committee be selected so that
it contains the married couple?
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• Solution
• If the committee contains the married couple,
  then only 2 men and 3 women are to be selected
  from the remaining 5 men and 8 women.
• The number of ways of selecting 2 men out of 5
  C52 10.
• The number of ways of selecting 3 women out of 8
  C83 56.
• the number of ways of selecting the committee
  lO 56 560.

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Example 14Find the number of ways a team of 4
can be chosen from 15 boys and 10 girls if (a)
it must contain 2 boys and 2 girls,

• Solution (a)
• Boys can be chosen in C152 105 ways
• Girls can be chosen in C102 45 ways.
• Total number of ways is 105 45 4725.

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(b) it must contain at least 1 boy and 1 girl.

• Solution
• If the team must contain at least 1 boy and 1
  girl it can be formed in the following ways
• (I) 1 boy and 3 girls, with C151 C103 1800
  ways,
• (ii) 2 boys and 2 girls, with 4725 ways,
• (iii) 3 boys and 1 girl, with C153 C101 4550
  ways.
• the total number of teams is
• 1800 4725 4550 11075.

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Example 15A woman has 12 friends and wishes to
invite 6 of them to a party. Find the numberof
ways she may do this if(a) there is no
restriction on choice,

• Solution (a)
• An unrestricted choice of 6 out of 12 gives
  C126 924.

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two of the friends is a couple and will not
attend separately,

• Solution(b)
• If the couple attend, the remaining 4 may then be
  chosen from the other 10 in C104 ways.
• If the couple does not attend, the woman simply
  chooses 6 from the other 10 in C106 ways.
• total number of ways is C104 C106 420.

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two of the friends are not speaking and will not
attend together

• Solution
• Let the non-speaking friends be A, B.
• If one of A, B attends, the party can be formed
  in C21 C105 ways.
• If neither A nor B attends, the party can be
  formed in C106 ways.
• Total number of ways is
• C21 C105 C106 714.

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Example 16Find the number of ways in which 30
students can be divided into three groups, each
of 10 students, if the order of the groups and
the arrangement of the students in a group are
immaterial.
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• Solution
• Let the groups be denoted by A, B and C. Since
  the arrangement of the students in a group is
  immaterial,
• group A can be selected from the 30 students in
  C3010 ways .
• Group B can be selected from the remaining 20
  students in C2010 ways.
• There is only 1 way of forming group C from the
  remaining 10 students.

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• Since the order of the groups is immaterial, we
  have to divide the product C3010 C2010 C1010
  by 3!,
• hence the total number of ways of forming the
  three groups is

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Example 17What is the greatest possible number
of points of intersection of 6 lines and 5
circles?
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• (i) Consider the 6 lines, the greatest possible
• number of points of intersection is C62
  15.
• (ii) Consider the 5 circles, the greatest
  possible
• number of points of intersection is 2 C52
  20.
• (iii) Each line can intersect a circle at 0, 1 or
  2
• points. Hence the greatest possible number
  of
• points of intersection resulting from the
• intersection of a line and a circle is
• 2 6 5 60.
• Combining (i)(ii)(iii), the greatest
  possible
• number of points of intersection is
  152060
• 
  95.

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Class Practice 1.7 P17

• (1)
• In how many ways can a team of three boys and
  three girls be chosen from six boys and seven
  girls ?

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Class Practice 1.7 P17

• (2)
• In how many ways can a selection of 4 letters be
  taken from the letters in the word COMPUTE if
• (a) the letter P is not to be chosen,
• (b) the letter P and E cannot be chosen in the
  same selection ?

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Class Practice 1.7 P17

• (3)
• A committee of 5 is to be chosen from a group of
  10 people.
• In how many ways can the committee be formed, if
  two particular people agree to serve only if they
  are both chosen?

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Class Practice 1.7 P18

• (4)
• A sorority house has 3 bedrooms and 10 students.
• One bedroom has 5 beds, the second has 3 beds,
  and the third has 2 beds.
• In how many different ways can the students be
  assigned rooms?

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Class Practice 1.7 P18

• (5)
• In how many ways can 12 students be divided into
  three groups of 4,
• if the order of the groups and the arrangement of
  the students in a group are immaterial ?

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Class Practice 1.7 P18

• (6)
• Eight policemen are to be posted to guard three
  separate buildings.
• In how many ways may they all be posted
• if no building is to be guarded by less than
  two policemen?

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Class Practice 1.7 P18

• (7) Code numbers, each containing three digits,
• are to be formed from the nine digits 1,
  2,
• 3,,9. In any number no particular digit
• may occur more than once.
• (a) How many different code numbers may be
• formed, and in how many of these will 9
  be
• one of the three digits selected?
• (b) In how many numbers will the three digits
• occur in their natural order (i.e. the digits
  being in ascending order of magnitude reading
  from left to right, e.g. 238) ?

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Ex 1b P19

• 1.(a) A firm has 12 computer
• programmers. Three of these people are
• to be promoted to system analysts.
• In how many ways can the 3 people to
• be promoted be selected?
• (b) A new product team will contain three
• of eight engineers, two of five
  marketing
• specialists, and one of three financial
• experts.
• How many different teams are possible?

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Ex 1b P19

• (2)
• Of the first 10 questions on a test, a student
  must answer 7.
• On the second 5 questions, he must answer 3.
• In how many ways can this be done?

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Ex 1b P19

• (3)
• Find the number of points of intersection of 15
  straight lines ,
• no two of which are parallel and no three of
  which are concurrent.

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Ex 1b P19

• (4)
• How many line segments are determined by the 5
  vertices of a pentagon ?
• Of these, how many are diagonals ?

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Ex 1b P19

• (5)
• There are 8 points on a circle.
• How many triangles can be inscribed with these
  points as vertices?

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Ex 1b P19

• (6)
• Given 10 points in a plane, no 3 of which are
  collinear and no 4 of which are concyclic,
• find the number of circles which may be drawn to
  pass through 3 of the given points.

P19
107
Ex 1b P19

• 7.(a) Find the number of different
• permutations of the letters of the
  word
• PROBABILITY
• (b) Find the number of different selections of
• 5 letters which can be made from the
• letters of the word PROBABILITY.
• (c) Find the number of ways in which
• (i) a selection,
• (ii) an arrangement can be made of 4
• letters taken from the letters
  of the
• word ARRANGE.

P19
108
Ex 1b P19

• (8)
• A football team consisting of a goalkeeper and 10
  other players is to be selected from 18 players.
• Just 2 of the 18 players are goalkeepers.
• Find the number of ways in which the team may be
  selected.

P19
109
Ex 1b P19

• (9)
• (a) In how many ways can 9 men
• be divided into three groups
• of 2, 3 and 4 respectively?
• (b) In how many ways can 9 men
• be divided into three groups of
• three if no regard is paid to the
• order of the group?

P19
110
Ex 1b P20

• (10)
• A committee of 4 is selected from a group of 8
  boys and 6 girls.
• If the committee must have at least one girl, in
  how many ways can the committee be selected ?

P20
111
Ex 1b P20

• (11)
• Of 20 computer chips, 4 will be selected for
  testing. How many different samples could be
  selected ?
• Suppose 5 of the 20 chips are defective and 15 of
  the chip are good.
• (a) How many of the samples contain only
• good chips?
• (b) How many of the samples contain 2 good
• chips and 2 defective chips?
• (c) How many of the samples contain one or
• more defective chips?

P20
112
Ex 1b P20

• (12)
• Two lines are parallel. On the first line there
  are 5 dots, and on the second there are 4.
• How many possible triangles can be formed by
  joining 3 of these dots?

P20
113
Ex 1b P20

• (13)
• (a) In how many ways can 10
• basketball players be divided into
• two teams of 5 players each?
• (b) In how many ways can 10
• basketball players be divided into
• two teams of 5 players so that the
• 2 best players are on opposite
• teams?

P20
114
Ex 1b P20

• (14)
• From 8 persons, including Mr. and Mrs. Chan, a
  committee of four is to be chosen.
• Mrs. Chan will not join the committee without her
  husband, but Mr. Chan will join the committee
  without his wife.
• In how many ways can the committee be formed?

P20
115
Ex 1b P20

• (15)
• A party of nine person is to travel in two cars,
  one of which will hold not more than seven
  persons, and the other not more than four.
• In how many ways can the party travel ?

P20
116
Ex 1b P20

• (16)
• In how many ways can three different numbers be
  selected from the thirty numbers 1, 2,., 30 such
  that their sum is
• (a) divisible by 2,
• (b) divisible by 3 ?

P20
117
Ex 1b P20

• (17)
• Two straight lines intersect at 0. If A1, A2,
  ...., An , are taken on one line, and B1, B2,
  ...., Bn on the other.
• Prove that the number of triangles that can be
  drawn with three of the points for vertices is
• (a) n2 (n 1) if the point 0 cannot be
• used,(b) n3 if 0 may be used.

P20
118
Ex 1b P21

• (18)
• A committee of 3 is to be chosen from 4 married
  couples.
• Find how many ways can the committee be chosen if
• (a) the committee must consist of one
• woman and two men,
• (b) all are eligible except that a
• husband and wife cannot both
• serve on the committee.

P21
119
Ex 1b P21

• (19)
• A table-tennis club is to select a team of three
  pairs, each pair consisting of a man and a woman,
  for a match in mixed double.
• The team is to be chosen from 7 men and 5 women.
• In how many different ways can the three pairs be
  chosen ?

P21
120
Ex 1b P21

• (20)
• There are 10 articles, 2 of which are alike and
  the rest all different.
• In how many ways can a selection of 5 articles be
  made ?

P21