Download Notes and Exercises

COURSE:
STATISTICS: PERMUTATIONS & COMBINATIONS
INSTRUCTOR: DOMINIC S. ASIEDU (MR.)
INTRODUCTION:
Introduction: When a choice is made disregarding the order of selection, it is called
combination and when the order is important, it is called a permutation.
PERMUTATIONS:
A permutation is an ordered arrangement of items or objects.
Permutation of n different objects
The number of permutations of n different objects taken r at a time is written as Pr or
nP r
, where : =
n
n!
Pr = (n – r)!
Therefore the number of permutations of n different objects taken n (i.e. all) at a time is
n
n!
Pr = (n – n)!
=
n!
0! = n! since 0! = 1
When applying the formula for n Pr in questions, we must make sure that:
1. The choice to be made is permutation and not a combination
2. Once an item is chosen it cannot be repeated. For example, if we have to form a number of 3
digits using digits 1, 2, 3, 4, 5 then, if repletion of digits is allowed, we cannot us the formula
of n Pr .
QUESTIONS 1:
i.
ii.
iii.
iv.
In how many ways can the letters A, B, C be arranged in a row?
In how many ways can the letters A, B, C be arranged taken two at a time?
In how many ways can the letters MATHS be arranged?
In how many ways can the letters MEASURING be arranged?
QUESTIONS 2:
i.
ii.
iii.
In how many ways can 10 people be seated on a bench, if only 4 seats are available?
In how many ways can 5 different objects be arranged taken 2 at a time?
In the managing of a committee of a society, there are nine members. In how many ways
can a chairman, a vice- chairman and a treasurer be selected from amongst them, if a
person is eligible for one post only?
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Permutation of with identical objects
i. The number of permutations of n objects taken all at a time, when p of the objects are
identical and the rest are all different is given by:
n!
P!
For example, the number of ways the letters MIME can be arranged is
4!
2!
= 4 x 3 = 12
Note that here there are 4 letters, so n = 4 and 2 Ms, so p = 2
ii.
The number of permutations of n objects taken all at a time, when p of the objects are
identical and of kind and q of them are identical and of a second kind and the rest are all
different is given by:
n!
p!q!
In general, in a set of n objects, when p of them are identical, q of them are identical and r of
them are identical, the number of permutations of these n objects taken all at a time is given by:
n!
p!q!r!
QUESTION 3:
i. In how many ways can the letters of the word MAXIMUM be arranged, when all are
taken at a time?
ii. In how many ways can the letters of the word STATISTICS be arranged, when all are
taken at a time?
iii. In how many ways can the letters of the word PARALLEL be arranged, when all are taken
at a time?
QUESTION 4:
i. In how many ways can the letters of the word SYNTHESIS be arranged, when all are
taken at a time?
ii. In how many ways can the letters of the word OPTIMIZATION be arranged?
iii. How many different arrangement of the letter of the word COMMISSION are possible?
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Conditional Permutations
In certain permutations, the possible arrangements of the objects are restricted in some way or
there are conditions imposed on some of the objects as in the following questions:
QUESTION 5:
i. In how many ways can the letters of the word LETTER be arranged amongst themselves?
In how many arrangements will the two T’s be together?
ii. How many different arrangements of letters can be made by using all the letters of the
word MINIMUM? In how many of these are vowels separated?
QUESTION 6:
i. Find the number of ways in which letters of the word SHALLOW can be arranged if
The two Ls must always be together;
The two Ls must not be together.
ii. How many different arrangements of letters of the word BEGIN are there which start with
a vowel?
iii. Find the number of ways in which letters of the word SYLLABUS can be arranged if
a. The two Ls must always be together;
b. The two Ls must not be together.
Cyclic Permutations
1. The number of ways r objects chosen from n different objects can be arranged in a circle
is:
n!
(n – r)!r
2. The number ways r objects chosen from n different objects can be arranged in a ring is:
n!
2(n – r)! r
3. If r = n i.e. if all n different objects are arranged in a circle then the number of ways is:
n! = (n – 1)!
n
4. If the objects are fixed on a ring, which can be turned over, the number of arrangements
will be:
n!
2n
(n –=1)!
2
QUESTION 7:
i. The representatives of five countries attend a conference. In how many ways can they be
seated at a round table?
ii. In how many ways can six different colored beads be arranged in a ring?
iii. In how many ways can five beads, chosen from eight different beads be threaded on to a
ring?
a. In how many ways can six people be seated at a round table?
b. How many ways are there if two people must sit next to each other?
c. How many ways are there if two people must not sit next to each other?
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COMBINATIONS:
A combination is a selection of a number of objects when the order of arrangement is not
important. There is only combination of letters ABC.
The number of combinations of a set of n different objects, taken r at a time or the selection of r
objects out of n objects is written
C r or (nr ), where
n!
n
C r = ( nr n ) =
r ! (n – r)!
n
n
Thus the number of ways of choosing or selecting r objects from a total of n objects is
n
n
C r or ( r ). Note that
n
n
C r=
Pr
r!
n
Note that
4! = 4 x 3 x 2 x 1 = 24
3! = 3 x 2 x 1 = 6 and by definition 0! = 1.
8!
6!
n
= 8 x 7;
20!
17! = 20 x 19 x 18;
10!
= 10 etc.
9!
C r = n C n – r ; 5C 2 = 5C 3, 9C3 = 9C6 etc.
The student should at this know how to calculate combinations. A calculating machine or
mathematical tables can be used.
QUESTION 1:
i. In how many ways can a football team be selected from a class of 20 students?
ii. In how many ways can four boys be chosen from six boys?
iii. How many ways can 9 mangoes be selected from a basket containing twelve mangoes?
iv.
In how many ways can three equal prizes be given to five boys?
v. In how many ways can three vacancies be filled out by 5 applicants?
QUESTION 2:
i. In how many ways can a committee of 3 ladies and 2 men be selected from a group of 7
ladies and 5 men?
ii. A committee of 1 girl and 2 boys is to be formed out of 5 girls and 6 boys. In how many
ways can this be done?
iii. In how many ways can two red balls and three black balls be chosen from a bag
containing five red and six black balls?
iv.
Five odd and four even numbers are to be selected from the numbers 1, 2, 3, 4, …., 15.
How many ways can this be done?
v. In how many ways can a committee of 6 men and 4 women be formed from a group of 8
men and 7 women?
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vi. Simplify (
If
3n
n+1)
3
- ( n -1 )
3
C 2 = 15 find the value of n
Drawing with replacement versus drawing without replacement
Supposing in a bag containing 3 balls, you take a ball at random from the bag and then replace it,
the result of the second draw does not depend on the ball you drew the first time. If you label the
balls A, B and C then possible draws are shown below.
AA, AB, AC, BB, BC, CC
i.e. we have 6 ways of drawing 2 balls one after the other with replacement out of a total of 3.
However, if the balls are drawn at random from the box one after the other without
replacement, drawing of the second ball is dependent on the first draw. In such a case, the
possible draws will be
AB, AC, BC, i.e. 3 ways = 3 C 2 ways
Again, if we have 4 balls labeled A, B, C and D and we want to draw two balls one after the
other with replacement, then the possible draws will be:
AA, AB, AC, AD, BB, BC, BD, CC, CD, DD
i.e. 10 ways
On the other hand, if the two balls are drawn at random from the box on after the other without
replacement,
AB, AC, AD, BC, BD, CD
i.e. 6 ways = 4 C 2 ways
We can therefore apply combinations to determine the number of ways of drawing objects from
a collection without replacement.
QUESTION 3:
1. How many ways can 4 red and 2 blue balls be drawn (without replacement) from a bag of
6 red and 4 blue balls?
Conditional Combinations
QUESTION 4:
i. A sub committee of 6, including a chairperson, is to be chosen from a main committee of
12. If the chairperson is to be a specified member of the main committee, in how many
ways can the sub committee be chosen?
ii. There are 15 boys and 10 girls, out of whom a committee of 3 boys and 2 girls is to be
formed. Find the number of ways in which this can be done if:
There is nor restriction
A particular boy is included
A particular girl is excluded
iii. In how many ways can five boys be chosen from a class of twenty boys if the class
captain has to be included?
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iv.
Out of 5 mathematicians and 7 physicists a committee consisting of 2 mathematicians
and 3 physicists is to be formed. In how many ways can this be done if
a) Any mathematician and any physicist can be selected;
b) Two particular mathematicians cannot be selected;
c) One particular Physicist must be selected.
QUESTION 5:
i.
ii.
iii.
iv.
A committee of person is to be formed from 5 men and 4 women, find the number of
ways of forming the committee if;
a. There must be only two women;
b. One particular woman must be included.
A committee of 3 is to be formed from 5 men and 3 women. Find:
a. The number of ways of forming the committee;
b. The number of ways of choosing at least one woman on the committee.
A committee of two men and three women is to be chosen from five men and four
women.
a. Find the number of ways of forming the committee;
b. If one of the women refuses to serve on the same committee with a particular
man, how many ways can the committee be formed?
Find the number of ways of forming a committee of 4 people can be chosen from a group
of 5 men and 7 women when it contains:
a. Only people of the same sex;
b. People of both sexes and there are at least as many women as men.
QUESTION 6:
i.
A committee of four is chosen from five teachers and three SS3 students. In how many
ways can this be done so that the committee contains:
a. At least one teacher
b. At least one teacher and SS3 student.
ii.
A company which has 10 directors has to send 3 of them to a conference.
a. In how many different ways can the 3 directors who attend be chosen?
b. It is now decided that at least one of those attending should be able to speak
French. Four out of 10 directors speak French. In how many ways can the 3
directors now be chosen?
iii.
Find the number of ways a committee of 4 people can be chosen from 6 boys and 6 girls
if it must contains:
a. 2 boys and 2 girls;
b. At least 1 boy and 1 girl.
iv.
A bag contains two white and three red balls. In how many ways can three balls be
chosen if:
c. At least one ball must be white;
d. At least one ball must be red.
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COURSE:
MODULAR ARITHMETIC / BINARY OPERATIONS
INSTRUCTOR: DAUDA ANDREW CONTEH (MR.)
PART I:
QUESTIONS
1. Find the least positive integer of x in the following:
a.
e.
(6 x 5) = 3 mod x
(
3 = 2 mod x
b.
f.
5 (mod 7) = x
2
x + 1 = 3 mod 7
c.
g.
3
23 + 14) = x (mod 5)
h.
2
x + 3 = 1 (mod 6)
x = 1 mod 5
d.
x + 1 = 4 (mod 7)
2.
a. Draw the:
i.
+
ii.
tables for the x = {1, 2,+3, 4, 5, 6} module 6
Addition
Multiplication
b. From your tables, find the truth set of:
i.
(4 n) + n = 4
ii.
t +(t
3) = 2
+
+
3.
a. Copy and complete the multiplication table modulo 5 on the set {1, 2, 3, 4}
1
2
3
4
1
2
1 +
4
3
4
2
3
b. From your table, solve the expression 2n
4≡3
+ n) = 2
c. Find the truth set of 2
(3
d. Find the truth set of m such that+(m + m) = 4
+
4. a. Construct the:
a. Addition
+ tables for the set H = {1, 5, 8, 12} mod 13
b. Multiplication
b. Use the tables to find the truth set of:
i.
n
n = 12
+ =5
ii.
n (8 (x) n)
+
a. Evaluate:
(12 12)
(8
8)
+
+
+
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COURSE:
MODULAR ARITHMETIC / BINARY OPERATIONS
INSTRUCTOR: DAUDA ANDREW CONTEH (MR.)
PART II:
BINARY OPERATIONS
A binary operation * on a non – empty set R of real numbers is a rule which combines any two
numbers a and b ΣR to produce another number, if the operations is not one of (+), (-), (x), (/)
(U), (∩), then the operation will be defined for you.
1. A binary operation * is defined on the set
S = {2, 3, 5, 7} by x * y = x + y -2 where x, y ΣS.
a) Copy and complete the table below:
*
2
3
5
7
2
2
3
5
5
7
4
8
8
10
b)
i.
Show whether or not S is closed under *
ii. Evaluate 3 * (2 * 5)
Note: If a * b produces a number which belongs to the set R, for all a, b ΣR, then set R is
closed under the operation.
2. A binary operation * is defined on R, the set of real numbers, by
x*y=
xy
+ x + y for all x, y ΣR.
4
Find:
i.
The identity element ℮ ΣR,
ii. The inverse of an element r ΣR
iii. Solve the equation x * 6 = 36
(WASSCE NOV. 2003)
Note: If ℮ ΣR is the identity or neutral element under a binary operation * on a set R of
real number, then; a * ℮ = ℮ * a; for a ΣR
That is when we combine any number with the identity element under a given operation
we get the number.
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3. An operation
is defined over the set R of real numbers by P
Q = P + Q + PQ
a) Show that
is commutative
b) Find the identity element for the operation
c) Find the inverse (under ) of the real number P, stating the value of P for which no
inverse exists.
d) Find the inverse -2
4. An operation * is defined over the set R of Real numbers by p * q = p + q + pq
a. Determine whether or not the operation is commutative
b. Find the neutral element ℮, under the operation
c. Determine whether or not, to each real number x, there corresponds a real number y,
such that q * p = ℮
d. Find q such that (p * q) * p = 3 stating the values of p for which there is no such q.
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COURSE:
GEOMETRICAL CONSTRUCTION (LOCI)
INSTRUCTOR: LAWRENCE OSEMEKHAI (MR.)
QUESTION 1.
a.
Using a ruler and a pair of compasses only, construct:
i.
ii.
iii.
b.
c.
d.
A triangle PQR such that PQ = 10cm QR = 7cm and PQR = 900;
The locus L1 of points equidistant from Q and R;
The locus L2 of points equidistant from P and Q.
Locate the point O equidistant from P, Q and R.
With O as Centre, draw the circum-circle of the triangle PQR.
Measure the radius of the circum circle.
QUESTION 2.
a.
Using ruler and pair
of compasses only, construct:
i.
A quadrilateral PQRS such that PQ = 7cm
1350 and QS = QR ;
ii.
iii.
QPS = 600, PS = 6.5cm,
PQR =
The locus L1 of points equidistant from P and Q;
The locus L2 of points equidistant from P and S.
b.
i.
Label the point T
where L1 and L2 intersect.
ii.
With centre T and
radius TP , construct a circle L3.
QUESTION 3.
a. Using ruler and pair of compasses only, construct:
a) A quadrilateral PQRS such that PQ = 10cm QR = 8cm, PS = 6cm,
600 and
QPS = 750
b) The locus L1 of points equidistant from QR and RS;
c) The locus L2 of points equidistant from R and S.
b. Measure RS .
PQR =
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COURSE:
GEOMETRICAL CONSTRUCTION (LOCI)
INSTRUCTOR: LAWRENCE OSEMEKHAI (MR.)
QUESTION 4.
a.
Using ruler and pair of compasses only, construct:
i.
ii.
iii.
A triangle XYZ with XY = 8cm, YXZ = 600 and
The perpendicular ZT to meet XY in T;
Locus L1 of points equidistant from ZY and XY.
b.
XYZ = 300;
If L1 and ZT
intersect at S, measure ST .
QUESTION 5.
a. Using ruler and pair of compasses only,
a)
Construct a quadrilateral PXYQ such that PX = 9.9cm, QX = 10.2cm,
0
= 75 ; QY = 10.4 cm and PQ // XY.
b)
Construct the
i.
ii.
c)
d)
QPX
Locus L1 of points equidistant from X and Y;
Locus L2 of points equidistant from QY and YX;
Locate M, the point of intersection of L1 and L2.
Measure PM .
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YCF-MOBSE-CEMC- 2013
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