Download MATH 236: TEST 1 Solutions

University of KwaZulu-Natal, Pietermaritzburg Campus
School of Mathematical Sciences
MATH 236: TEST 1 Solutions
Time Allowed: 90 minutes
Instructions: Answer all the questions and show all working.
Total marks: 50
Total questions: 6
Bonus marks: 5
Question 1. [Total: 10]
(1) State without proof the Addition Principle
(5 points)
(2) Let S and T be finite sets. Using Addition Principle to prove that |S × T | = |S| · |T |. (5 points)
Solution
(1) If {A1 , A2 , · · · , Ak } is a pairwise disjoint collection of finite sets, then
| ∪ki=1 Ai | =
k
X
|Ai |.
i=1
(2) Let S = {x1 , x2 , · · · , xs } and T = {y1 , y2 , · · · , yt }, where s = |S| and t = |T |. Let Ai be the set
of ordered pairs whose first element is xi , that is,
Ai = {(xi , y) : y ∈ T }.
Clearly,
S × T = ∪si=1 Ai and |Ai | = t for 1 ≤ i ≤ s.
Also, {A1 , A2 , · · · , As } is a pairwise disjoint collection of subsets. Thus by Addition Principle,
we have
s
X
|S × T | = |A1 ∪ A2 ∪ · · · ∪ As | =
|Ai | = st = |S| · |T |.
i=1
Question 2. [Total: 10] Let R be a relation defined on the set S = {4, 5, 6, 11, 12, 13, 18} by xRy if and
only if 7 | 3x + 4y.
(1) Show that R is an equivalence relation.
(6 points)
(2) Find [4], the equivalence class containing 4.
(2 points)
(3) Find |S/R|
(2 points)
Solution.
(1) For any x ∈ S, we have that 3x + 4x = 7x is divisible by 7 so xRx, so R is reflexive.
Assume that xRy where x, y ∈ S. Then 7 | 3x + 4y. Since 7(x + y) − (3x + 4y) = 4x + 3y and
both 7(x + y) and 3x + 4y are divisible by 7, we deduce that 7 | 3y + 4x, so yRx and hence R
is symmetric.
Finally, assume that xRy and yRz for some x, y, z ∈ S. Then 3x + 4y = 7s and 3y + 4z = 7t
for some s, t ∈ Z. It follows that 3x + 4y + 3y + 4z = 7(s + t) and hence 3x + 4z = 7(s + t − y)
is divisible by 7. Thus xRz and therefore R is transitive.
Since R is reflexive, symmetric and transitive, it is an equivalence relation.
(2) We have
[4] = {y ∈ S : 4Ry} = {y ∈ S : 7 | 3 · 4 + 4y} = {4, 11, 18}.
2
(3) We have that
[5] = {5, 12}, [6] = {6, 13}
and
S = [4] ∪ [5] ∪ [6].
So |S/R| = 3.
Question 3. [Total: 10]
(1) How many ways can the letters of the word ECCEDENTESI be arranged in a
letters TESI must remain together as a block?
(2) How many ways can the committee of 6 be chosen from a group of 15 people if the
consists one president, two secretaries and 3 members?
row if the
(5 points)
committee
(5 points)
Solution.
(1) Consider ‘TESI’ as one block, there are 8 total letters, in which there are 3 letters ‘E’, 2 letters
‘C’, one letter ‘D’, on e letter ‘N’ and one letter ‘TESI’. Using the formula for permutations of
indistinguishable objects, there are exactly
8!
= 3360
P (8; 3, 2, 1, 1, 1) =
3! · 2! · 1! · 1! · 1!
ways.
(2) There are (15)1 ways
to choose a president from a group of 15 people; among 15 − 1 = 14 people
left, there are 14
2 ways to choose 2 secretaries and finally among 14 − 2 = 12 people, there are
12
3 ways to choose 3 committee members. Thus by the multiplication principle, there are
14
12
(15)1 ·
·
= 300300
2
3
ways.
Question 4. [Total: 10]
(1) How many odd integers are there from 100 to 999 ?
(5 points)
(2) How many solutions does the equation x1 + x2 + x3 = 9 have, where x1 , x2 , x3 are nonnegative
integers?
(5 points)
Solution.
(1) Every odd number from 100 to 999 has the form abc where a, b, c ∈ {0, 1, · · · , 9} and a 6= 0.
Since abc is odd, there are only 5 possibilities for c, since a 6= 0, there are only 9 possibilities
for a and finally, there are 10 possibilities for b. Thus by the multiplication principle, there are
9 · 10 · 5 = 450 odd numbers from 100 to 999.
(2) The numbers of the nonnegative solutions to the equation x1 + x2 + x3 = 9 is the number of
9-combinations from a set of size 3. Thus there are
3+9−1
11
11
=
=
= 55
9
9
2
solutions.
Question 5. [Total: 10]
(1) Find the coefficient of x6 y 5 in the expansion of (2x2 − 3y)8
(5 points)
(2) How many numbers between 1 and 84 inclusive, are divisible by at least one of the prime numbers
2, 3 and 7?
(5 points)
Solution.
3
(1) Let X = 2x2 and Y = −3y. Then (2x2 3y)8 = (X + Y )8 . By Binomial Theorem, the coefficient
of X 3 Y 5 in the expansion of (X + Y )8 is 83 and since
X 3 Y 5 = (2x2 )3 (−3y)5 = 23 (−3)5 x6 y 5 ,
we deduce that the coefficient of x6 y 5 in the expansion of (2x2 − 3y)8 is
8
23 · (−3)5 ·
= −108864.
3
(2) Let S = {1, 2, · · · , 84}. Let S2 be a subset of S consisting of all numbers divisible by 2, S3 be a
subset of S consisting of all numbers divisible by 3, and finally S7 be a subset of S consisting of
all numbers divisible by 7.
We have
|S2 | = |{x ∈ S : 2 | x}|
= |{x ∈ S : x = 2k, 1 ≤ 2k ≤ 84}|
= |{x ∈ S : x = 2k, 1 ≤ k ≤ 84
2 }|
= |{x ∈ S : x = 2k, 1 ≤ k ≤ 42}|
= 42.
Similarly, we have
|S3 |
= 84
3 = 28
|S7 |
= 84
7 = 12
|S2 ∩ S3 |
= 84
6 = 14
|S2 ∩ S7 |
= 84
14 = 6
|S3 ∩ S7 |
= 84
21 = 4
|S2 ∩ S3 ∩ S7 | = 84
42 = 2.
By the Inclusion-Exclusion Principle, we have
|S2 ∪ S3 ∪ S7 | = |S2 | + |S3 | + |S7 | − |S2 ∩ S3 | − |S2 ∩ S7 | − |S3 ∩ S7 | + |S2 ∩ S3 ∩ S7 |
= 42 + 28 + 12 − 14 − 6 − 4 + 2
= 60.
Hence, there are exactly 60 numbers between 1 and 84 inclusive which are divisible by at least
one of the numbers 2, 3 and 7.
Extra Marks. Let n and k be integers with 1 ≤ k ≤ n. Prove the identity
n
n−1
k
=n
k
k−1
by using a combinatorial proof.
(5 points)
Solution.
Let S be a set consisting of n people. We want to count the number of ways we can choose group of
k-people and then choose
a leader from this group.
Clearly, there are nk groups of k-people from a set of size n. Among each group of k-people, there
are k1 = k ways to choose a leader from this group. Therefore, by the Multiplication Principle, there
are nk k1 = k nk ways.
We next count this number in a different way. We now choose a leader from the whole group of
n-people and then among
n − 1 people left, we choose a group of k − 1 people withthe chosen leader.
Obviously, there are n1 = n ways to choose a leader first and then there are n−1
ways to choose a
k−1
n−1
group of (k − 1)- people. Thus by the Multiplication Principle, there are n1 n−1
=
n
k−1
k−1 ways.
Now by the counting principle, we obtain that
n
n−1
k
=n
.
k
k−1