2021-22 First Semester
MATH 3600 Discrete Mathematics
Test 1
Report
Overall performance statistics
Score range
[10, 20) [20, 30) [30, 40) [40, 50) [50, 60) [60, 70) [70, 80) [80, 90)
Frequency
3
4
2
1
6
3
2
3
Corresponding grade
F
F
D
C
B-C
B
A
A
It should be noted that the corresponding grade is only a rough reference, as you get a single grade
for the whole course (which counts performance in the tests/exam as well as participation) rather
than a separate grade for each component. The thresholds for different grades also depend on a
range of factors such as the level of difficulty of the questions.
Item statistics
Question
1(a)
1(b)
1(c)
1(d)
1(e)
Mean percentage
70.8
95.8
37.5
25.0
4.2
Question
2(a)
2(b)
2(c)
2(d)
2(e)
Mean percentage
79.2
33.3
29.2
12.5
54.2
Question
3(a)
3(b)
3(c)
Mean percentage
59.7
60.4
72.9
Question
4(a)
4(b)
4(c)
Mean percentage
69.4
75.0
28.0
Question
5
Mean percentage
28.3
Overall comments
The overall performance is diverse. Some results are impressive given that there are some challenging
questions in the paper, but at the same time there are also some easy and standard questions so
some scores on the other end are far from being satisfactory. As mentioned before, no matter you are
aiming for a pass or for a top grade, you should ensure that you don’t miss the easy and standard
questions (such as Questions 1(a), 1(b) and 3). With sufficient preparation, these are the questions
which everybody should be able to answer correctly (not to mention also that you are allowed to
bring a piece of cheat sheet to the test).
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Discussions on the questions
1. (Scoring: 4 marks for each part)
(a) This is straightforward
and yet almost one-third of the students missed it. The answer is
−5
simply
· 46 .
6
12
(b) This is straightforward and almost everybody got it right. The answer is simply
=
4, 0, 3, 5
12!
.
4!3!5!
(c) This part is supposed to be an easy application of the ‘n choose r table’ from Chapter 2,
and the result is simply H34 · 47 . Yet less than half of the students got the correct answer.
(d) This is just Question 5(a) in Discussion Session 5, albeit with 1000 dollars changed to
1500 so now you have to consider two cases when computing the coefficient of x1500 in the
generating function. Only a quarter of the students got it right.
(e) This part is challenging. There are many different ways to count, but apparently none of
them is straightforward and all would involve consideration of multiple cases. One of the
easiest solutions is probably to count the opposite — the number of choices which do not
form a triangle. Suppose rods of lengths a, b, c are chosen, where 1 ≤ a < b < c ≤ 50.
These rods do not form a triangle if and only if c ≥ a + b. Suppose a + b = S. Then S
may range from 3 to 50 (if S > 50 there is no choice for c). For each S, there are 51 − S
choices for c (from S to 50). Also, there are b 12 (S − 1)c choices for (a, b) — for instance
if S = 10 there are 4 choices for (a, b), namely, (1, 9), (2, 8), (3, 7) and (4, 6). Hence the
number of choices which do not form a triangle is
50 X
S−1
S=3
2
· (51 − S) =
24
X
k=1
k(50 − 2k) +
24
X
k(49 − 2k)
k=1
where in the last step we separate into odd and even S (by setting S = 2k + 1 and
S = 2k + 2 respectively). The above sum is not hard to compute (just summing 99k − 4k 2
from 1 to 24), and subtracting it from 50
gives the answer.
3
2. (Scoring: 4 marks for each part)
20
(a) The performance in this part was good. All you need to note is that (1+3)
20 X
20 k
=
3
k
k=0
by the binomial theorem.
(b) The technique of extracting terms of a certain parity has been encountered on several
occasions. If we let g(x) be the expression then the answer is simply 12 [g(1) − g(−1)].
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n+1
n+1
n+1
(c) This should be straightforward and the answer is simply 2
+4
+6
,
1
2
3
but less than one-third of the students got it right. Some students gave a quadratic
polynomial as the answer. Apparently these students tried to find hn instead of h0 + h1 +
· · · + hn .
(d) A number of students tried to invoke the number Qn from Chapter 6. That does not
sound right here because Qn counts the number of permutations with certain properties,
while the problem here is not quite related to permutations. Essentially this problem is
‘choosing 10 out of 7201 items’ where no consecutive items are chosen, so it is equivalent
to first write down 7191 copies of ‘X’ (corresponding to items not to be chosen) and then
insert 10 copies of ‘O’ (corresponding to items to be chosen) in the 7192 possible positions:
↓
↓
X
↓
↓
X
X
↓
···
↓
X
↓
X
(e) Essentially we have to derange the 50 odd numbers and the 49 even numbers separately.
Some students miscounted the number of odd and even numbers in the set {1, 2, ..., 99}.
3. (Scoring: 6 marks for (a), 6 marks for (b) and 8 marks for (c))
This question is supposed to be easy as it consists of entirely bookwork-type computations.
Some common mistakes are listed below.
(a) The coefficient of z n in (1 − z)−1 should be z n instead of (−1)n z n . Also some students
tried to expand (x − 2)−1 directly which led to some mess.
(b) The coefficient of xn in the expansion of (1 − x)−5/2 should be (−1)n −5/2
instead of
n
−5/2
.
n
(c) Some students tried a particular solution of the form p · 3n , which would not work since
this is a solution to the homogeneous part.
4. (Scoring: 6 marks for (a), 7 marks for (b) and 7 marks for (c))
This is a typical question on giving combinatorial proofs to algebraic equalities. There are
generally two ways to do this — counting the same thing in two ways to show that two different
expressions are equal, and to establish a bijection between two sets to show that they have the
same size. It should be noted that using algebraic manipulations to argue that two quantities
are equal (or to substantially supplement a combinatorial proof) is not acceptable.
In this question all three parts can be proved by counting in two ways. Indeed both (a) and
(c) arose during the discussion sessions, so you should have seen them before.
When giving such a combinatorial proof, you need to justify that you are counting the same
thing in two ways (some students just said that the LHS counts something and the RHS also
counts something, but this does not prove anything unless the two things being counted are the
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same). Furthermore, You need to state the conditions of the counting problem clearly. Using
(a) as an example, some students just claimed that the LHS is the number of ways to put 3
balls into n boxes. There are important questions such as
• whether the balls distinct or identical;
• whether the boxes distinct or identical;
• whether there is any restriction (e.g. can two balls go into the same box?),
and these must be clarified.
5. This question is challenging. The root can of course be easily expressed in square root form
using the quadratic formula, and while expanding the square root using Newton’s binomial
theorem, it is a good idea to rewrite the expression inside the square root into the form 1 + z.
Knowing the expression for 1/2
should enable you to get an expression that looks like Catalan
k
numbers. However quite a number of students missed the final step of rigorously showing what
the question asks, namely, that the next 7 digits are 0000042. This would require considering
the residual terms and establish an upper bound. To see why this is important, note that
1
= 0.010204081632...
98
and while you can see a clear pattern in the digits (consisting of powers of 2), the next two
digits are not 64! (Why not?) As an exercise, how about
1
= 0.001002004008016032...?
998
Are the next 3 three digits 064? Can you justify?
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