Common Errors in Counting Problems
by
Scott Annin and Kevin Lai
Mathematics teachers are often asked: What is the most difficult topic for you to teach?
Our answer is easy: it's teaching students to count. Although the computations in combinatorics
are not necessarily difficult, the concepts can be challenging. Many times, no rigid procedures or
formulas can be directly employed to solve the problems, and students simply do not know
where and how to begin approaching these problems. Furthermore, subtle differences in
questions posed can necessitate entirely different solution techniques.
Combinatorics is an important topic in the mathematics curriculum. For example, Kapur
(1970) observed that it can help students develop their reasoning skills (making conjectures,
generalizing, systematic thinking) and conceptual understanding of mapping, functions,
equivalence relations, etc. Combinatorics also has applications in many fields such as physics
and biology, and Kapur recommended that it should be taught at all grades. The same sentiment
is expressed by Principles and Standards for School Mathematics (NCTM 2000), which observes
that although the topic of Discrete Mathematics was removed from the 2000 standards, the topic
is actually evident across all the other standards and spans from grades K-12. It recognizes the
wide usage of this area of mathematics in business and industry, and encourages students to
develop their understanding of combinatorics throughout their school years.
A common strategy among Discrete Mathematics textbooks is to attempt to classify
counting problems neatly according to various types in the hopes of giving students some clear
guidelines and procedures (Brualdi, 2004; Johnsonbaugh, 2009; Rosen, 2007). This way of
classification is also reflected in the research literature (Fischbein & Gazit 1988; Batanero et al.
1992). In counting arrangements of objects, for example, one can distinguish problems
according to whether repetition of objects is allowed and whether the order that those objects are
placed in makes a difference to the arrangement. This results in four general types of counting
problems. Provided that students are studying problems that rightly belong in one and only one
of these four categories, there does not appear to be anything extraordinarily unusual about what
the student must do to successfully learn the material. Figure 1 shows the number of
arrangements of n objects in each of the four traditional cases.
Repetition Allowed
r-permutations
of n objects
r-combinations
of n objects
nr
Repetition Not Allowed
n!
 n  r !
(n  r  1)!
n!
r ! n  1!
r ! n  r !
Fig. 1: Common table seen in Discrete Mathematics textbooks
What we often find challenging about teaching students to count, however, is that most
problems do not fall cleanly into one and only one of the standard categories of counting
problems. Rather, each problem typically involves different aspects that rely on different
techniques. Moreover, subtle differences from one problem to the next lead to a propensity for
hesitation and uncertainty in the student's thinking. In this article, we will present a few
examples of such problems, illustrate their subtleties, and provide some guidance and practice
with this subject. Let us begin with a few problems:

Problem 1: How many flags of eight horizontal stripes contain at least six blue stripes, if
each stripe is colored red, green, or blue?

Problem 2: How many flags of eight horizontal stripes contain at least two blue stripes,
if each stripe is colored red, green, or blue?
Fig. 2: Examples for Problems 1 and 2

Problem 3: How many 4-card hands consisting of two cards each of two different kinds
can be made from a standard deck of 52 cards?

Problem 4: How many full house* card hands can be made from a standard deck of 52
cards?
Fig. 3: Examples for Problems 3 and 4
*A full house consists of precisely three cards of one kind and two cards of a second kind from a
standard deck.
All four of these problems involve aspects that must be handled carefully. Many students do not
see these delicate points, nor do they see any substantive differences in the way these four
problems are best handled. This paper is intended to draw attention to some of the errors in
thinking that students fall victim to in solving counting problems such as these, examine the
differences between the correct solutions to these problems and the solutions often given by
misdirected students, and provide some problems to encourage practice in this challenging area.
Over-Counting due to Order
In my many years of teaching combinatorics and asking exam questions like Problem 1, I (S.
Annin) have often found that students arrive at the following incorrect solution.
Incorrect Solution to Problem 1: Consider the flag shown below. Each stripe must be labeled
with a color: red, green, or blue. Since at least six blue stripes are required, one can begin by
choosing six stripes to color blue. There are C(8,6) = 28 ways to do this. Once the six blue
stripe minimum is fulfilled, the remaining two stripes (whichever ones they are) can each be
colored in any one of three ways (red, green, or blue). Therefore, there are 3*3 = 9 ways to
finish coloring the stripes, for a total of 28*9 = 252 ways to color the stripes of the flag.
Fig. 4: Flag Template for Problem 1
Analysis: The solution above appears to be a straightforward application of the Multiplication
Principle: to count the total number of ways of doing a task, one can multiply the number of
ways of performing the subtasks that comprise the whole task (hence the answer 28*9).
However, this answer is nearly two times too large. The main reason for this is that the solution
presented above imposes an ordering on the blue stripes (students are unaware of this), when the
order in which the stripes are painted is immaterial: the question is only interested in counting
finished flags. In particular, the solution above first chooses six blue stripes, and then (possibly)
chooses additional blue stripes. Therefore, the solution has ordered the blue stripes into two
“groups”: the first six blue stripes and any additional blue stripes after that.
To see the difficulty more explicitly, consider, for example, the flag colored with all eight stripes
blue. There is obviously only one such flag, yet the solution above counts it 28 times in the 252
total obtained! How is that possible? To see how, consider two different ways to complete the
coloring of the flag: (1) choose the top six stripes to be blue, then choose to color both of the last
two stripes blue also, and (2) choose the bottom six stripes to be blue, then choose to color both
of the first two stripes blue also. In either case, we arrive at the same “all blue” flag, but from
the point of view of the solution above, these flag-colorings were obtained differently and are
therefore counted as two different flags. In fact, there would be 28 different flags if we kept
track of which six of eight stripes were chosen first. For instance, suppose we place a gold star
next to each stripe that is chosen among the first six for blue color. For an “all blue” flag, there
are C(8,6) = 28 different patterns of gold stars that could occur on the stripes. The solution
above incorrectly counts each of these 28 outcomes as a different flag.
Correct Solution to Problem 1: While it is possible to do subtraction to correct for the overcounting that has occurred above, it is easier to simply take care to avoid over-counting in the
first place. One might observe, for example, that the trouble described in the analysis above does
not arise on flags whose last two stripes are colored red or green. The number of such flags
(flags with exactly six blue stripes) is precisely C(8,6)*2*2 = 28*4 = 112. Next, the number of
flags that have exactly seven blue stripes can be counted. There are C(8,7) ways to choose seven
stripes to color blue, and the last stripe must be either red or green. Hence, there are C(8,7)*2 =
8*2 = 16 such flags. Finally, there is one flag with all eight stripes blue. Using the Sum Rule,
the total number of flags with at least six blue stripes is 112 + 16 + 1 = 129.
8 
8 
   2 2     2 1
6
7
Exactly 6 blues
Exactly 7 blues
Exactly 8 blues
Fig. 5: Correct Solution to Problem 1
Conclusions: Many counting problems involve a “minimum” requirement (e.g. flags with at
least six blue stripes, license plates with at least six A’s, bags of fruit containing at least six
apples, etc.) Problems such as these are vulnerable to over-counting in the manner observed
above, but this can be remedied by replacing the minimum condition with several cases
containing an “exact” condition and summing the total count for each of these cases. Instead of
counting the number of flags with at least six blue stripes in one step, for example, we can count
the number of flags with exactly six stripes, exactly seven stripes, and exactly eight stripes
separately. Each of these three subcases are less prone to error, and as we have done above, the
results from each subcase can be added to yield the final result.
Correct Solution to Problem 2: This problem is closely related to Problem 1 and can be solved
the same way. Since the number of blue stripes can now be any value 2,3,4,5,6,7, or 8, we have
seven different cases to take care of. We ask: how many flags of 8 stripes containing exactly k
blue stripes? The answer is C(8,k)*28-k, obtained by first choosing k stripes to color blue, and
then deciding on either red or green color (two choices each) for each of the remaining 8-k
stripes. We must sum this value for all k=2,3,4,5,6,7,8. The total is C(8,2)*64 + C(8,3)*32 +
C(8,4)*16 + C(8,5)*8 + C(8,6)*4 + C(8,7)*2 + C(8,8)*1 = 1,792 + 1,792 + 1,120 + 448 + 112 +
16 + 1 = 5,281.
Although the final answer requires a lot of terms by this solution, students become very
comfortable with it, especially after seeing enough examples similar to Problems 1 and 2. Of
course, when the “minimum” requirement on the problem is rephrased as the complement of a
“maximum” requirement, a shorter answer can be found: How many flags with eight horizontal
stripes contain at most one blue stripe? Since there are 28 flags with no blue stripes and
C(8,1)*27 flags with exactly one stripe, there are a total of 28 + C(8,1)*27 = 256 + 1,024 = 1,280
flags that do not contain at least two blue stripes. Since there are a total of 38 = 6,561 flags with
eight horizontal stripes (each independently colored either red, green, or blue), there are precisely
6,561 – 1,280 = 5,281 that do contain at least two blue stripes.
Order Issues in Another Context
Perhaps no subject matter is more common in combinatorics (and probability) problems than
playing cards. Despite the heavy analysis that playing cards have received, they can elicit some
tricky aspects that are easily overlooked. The pitfall we are about to describe arises in the
context of many other counting problems as well, and as with many difficulties encountered in
basic Combinatorics, ordering of objects is at the core of the issue. In particular, there is a basic
difference in how Problems 3 and 4 (stated above) are solved, and it has everything to do with
order.
Solution to Problem 3: The tasks required to construct a 4-card hand consisting of two pairs can
be described as follows: (1) choose two kinds of cards (i.e. ace, king, queen, …, 3, 2) to build
pairs from, and (2) choose two suits for the two cards from each of the two kinds chosen in the
first step. There are C(13,2) ways to complete task (1), and there are C(4,2)*C(4,2) ways to
complete task (2). This gives us a total of C(13,2)*C(4,2)*C(4,2) = 78*6*6 = 2,808 such hands.
13   4   4 
         2808
 2   2  2
Ways to
choose 2 kinds
Ways to choose 2
cards from one kind
Ways to choose 2 cards
from another kind
Fig. 6: Correct Solution to Problem 3
Analysis: By simultaneously choosing the two kinds of cards that the pairs will be built from,
we avoid the common mistake of putting the two kinds of cards in a specified order. Faulty
reasoning here would suggest that there are 13 ways to choose the first kind, C(4,2) ways to
choose two of the four suits of that kind, 12 ways to choose the second kind, and C(4,2) ways to
choose two of the four suits of the second kind. Multiplication of the various choices available
here would give the answer 5,616, twice as large as it should be. This faulty reasoning, in which
the two kinds are placed in order (we have emphasized this point by referring to them as “first
kind” and “second kind” above) when they should not be, is extremely common.
Students often successfully solve Problem 3 merely because everyone’s experience with playing
cards is that the order of the cards does not matter. That is, cards can be played from a hand in
any desired order, so the “left-to-right” order that the cards appear in the hand makes no
difference. Therefore, most people are trained to use combinations (rather than permutations) to
solve card problems. A natural corollary of this, of course, is that in card problems where order
is in some way significant, students often falter. Only a minor change to Problem 3, for
example, leads to the more troublesome Problem 4.
Solution to Problem 4: This problem appears to be almost identical to Problem 3. Again, we
must (1) choose two kinds of cards, and (2) choose an appropriate number of cards of each kind.
In fact, if one were to try a solution technique similar to that given for Problem 3 above, the
expected answer might be C(13,2)*C(4,3)*C(4,2). But this answer is incorrect, because it does
not distinguish in any way between the two chosen kinds. In reality, however, one of the two
chosen kinds of cards must have three suits selected from it, while the other should have only
two. The correct answer in this case would therefore be:
13 
 4  4
   2        3744
2 
3   2
Ways to
choose 2 kinds
Ways to choose the
kind from which 3
cards are taken
Ways to
choose 3 cards
from first kind
Ways to choose
2 cards from
second kind
Fig. 7: Correct Solution to Problem 4
Alternatively, we could solve it as:
 4
 4
13     12     3744
3
 2
Ways to
choose a kind
for 3 cards
Ways to
choose 3 cards
of that kind
Ways to choose
another kind for 2
cards
Fig. 8: Alternative Solution to Problem 4
Ways to choose 2
cards of that kind
Conclusions: In Problem 3, the order of the two kinds used to create two pairs is unimportant,
while in Problem 4, we must differentiate between the kind that three cards are chosen from and
the kind that two cards are chosen from. This distinction appears to be the result of the fact that
the same number of cards of each kind are chosen in Problem 3, whereas this is not the case in
Problem 4. Here are some related problems that also demonstrate this subtle distinction in the
problems. Try to solve the two exercises here before the reading the solutions that follow.
Exercise 1: Alice, Bob, Candice, Douglas, Ellen, Freddie, George, Henrietta, and Irvin are
instructed to break into one group of size 2, one group of size 3, and one group of size 4. In how
many ways can they do this?
Exercise 2: Alice, Bob, Candice, Douglas, Ellen, and Freddie, George, Henrietta, and Irvin are
instructed to break into three groups of three. In how many ways can they do this?
To solve Exercise 1, there are C(9,2) ways to choose two people for the group of size 2. Next,
there remain 7 people, and hence C(7,3) ways to choose three people for the group of size 3.
Finally, the last four people form a group of 4 by default. The total is C(9,2)*C(7,3) = 36*35 =
1,260.
Exercise 2 is rather challenging for students if attempted immediately after solving Exercise 1.
In solving Exercise 2, many students will follow the same reasoning exactly to obtain the answer
C(9,3)*C(6,3) = 84*20 = 1,680. This answer however, is six times too large! To see why, let us
consider a smaller example first: In how many ways can Alice, Bob, Candice, and Douglas break
into two groups of two? One might be tempted to give the answer C(4,2) = 6, the number of
ways to choose two of the four individuals to form a group (with the other two then being paired
up by default thereafter). However, we can directly count only 3, not 6, pairs of two: Alice and
Bob in a pair, Alice and Candice in a pair, and Alice and Douglas in a pair. The answer 6 would
only be correct if we wanted to place the two groups chosen in order. Perhaps, for example, the
first two individuals chosen to form a group are also designated as officers for the whole group.
If we wanted to keep track of this additional information, then there would be 6 possible
outcomes.
Returning to Exercise 2, the answer 1,680 obtained above involves unintentionally putting the
three groups in order. This was acceptable for Exercise 1, but not for Exercise 2. For example,
the group formation ABC | DEF | GHI is being counted differently from the group formation
GHI | ABC | DEF (note that the order of the three letters comprising a group is immaterial here).
There are 3! = 6 ways to permute the same three groups of three (see Figure 9), and thus, each
way of grouping nine individuals into three groups of three is being counted six times.
Therefore, the correct answer here is 1,680/6 = 280.
(1) ABC DEF GHI
(4) DEF GHI ABC
(2) ABC GHI DEF
(5) GHI ABC DEF
(3) DEF ABC GHI
(6) GHI DEF ABC
Fig. 9: The same three groups of three occur six times when an order on groups is imposed
One of the most important questions to resolve in solving counting problems is: Does the order
of the objects occurring in an arrangement matter? As we have seen in this section, however,
this question may not be as black and white as it appears at first. Great care must be exercised.
Conclusions
In this article, we used examples in a variety of contexts to show how minor differences in
combinatorics problems can result in major misconceptions in a novice solver. While the
standard categorization of these problems into different types can ease this difficulty to some
extent, many problems do not belong to one and only one category. A variety of problems need
to be presented to students so that they can gain a better conceptual understanding of the
problems and become familiar with the common pitfalls that must be avoided.
Teachers can readily and specifically address the issues we have raised in this article by giving
students any number of problems like the ones we have discussed here. In so doing, they will
equip and train their students to be mindful of the potential errors that frequently arise in solving
these problems. To that end, we conclude by providing the reader with an even greater variety of
examples, similar to those we have discussed, that may be useful in the combinatorics classroom:
Homework #1: How many license plates consisting of 9 capital letters contain at least six M’s?
Homework #2: How many 6-card poker hands can be formed that contain three cards from each
of two different suits? How many 6-card poker hands can be formed that contain four cards
from one suit and two cards from another?
Homework #3: (Modification of 2002 American Invitational Math Exam-Version 2, Problem 9).
Let S = {1,2,3,4,5,6,7,8,9,10}. In how many ways can two disjoint, non-empty subsets of S be
formed?
References
Batanero, C., Godino, J. D., Navarro-Pelayo, V. “Effect of the implicit combinatorial model on
combinatorial reasoning in secondary school pupils.” Educational Studies in Mathematics
32 (1997): 181-199.
Brualdi, R.A. Introductory Combinatorics, Prentice Hall, Upper Saddle River, NJ, 2004.
Fishbein, E. and Gazit, A. “The combinatorial solving capacity in children and adolescents.”
Zentralblatt fuer Didaktitk der Mathematik 5 (1988), 193-198.
Kapur, J. N. “Combinatorial Analysis and School Mathematics,” Educational Studies in
Mathematics 3 (1970): 111-127.
Johnsonbaugh, R.J. Discrete Mathematics, Prentice Hall, Upper Saddle River, NJ, 2009.
National Council of Teachers of Mathematics (NCTM). Principles and Standards for School
Mathematics. Reston, VA: NCTM, 2000.
Rosen, K.H. Discrete Mathematics and Its Applications, McGraw-Hill, New York, NY, 2007.