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Counting Problems
1. How many circles in the plane contain at least three of the points (0, 0), (0, 1),
(0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)?
Solution: There are 93 = 84 three element subsets, all of which give rise
to a circle except those which are co- linear, of which there are 8. But some
circles get counted four times, namely those which contain four of the nine
points. So the problem is to count the four element subsets that belong to
a circle. There are six such sets that are vertices of a square, and four more
that are vertices of a non-square rectangle. Then there are four more that are
isomorphic with {(0, 0), (1, 0), (2, 1), (2, 2)}. For each of these 4 + 6 + 4 = 14
sets, we must subtract three circles that have been counted 4 times. Therefore
there are 84 − 8 − (3 · 14) = 34 such circles. Alternatively, note that there are
8 circles with centers in the region x > 0, y > 0, thinking now that the nine
points have been translated so that the middle one is the origin. Thus there
are 8 · 4 + 2(circles with center at the origin) = 34.
2. Nine points are distributed around a circle in such a way that when all 92 pairs
of points are connected by segments, no three of the segments are concurrent.
How many points of intersection are there inside the circle?
Solution: Each point of intersection is determined by a four-element subset
of the nine-element set, so there are 94 = 126 such points.
3. Two points are randomly and simultaneously selected from the 4 × 5 grid of
20 lattice points {(m, n)|1 ≤ m ≤ 5 and 1 ≤ n ≤ 4}. What is the probability
that the distance between them is rational?
Solution: There are 4
5
2
4
= 30 and
2 are 20
= 190
2
= 40 ‘horizontal’ pairs of points and 5
there are two pairs of points which are five units apart. There
pairs of points, so the probability of two points being a rational distance apart
is 72/190 = 36/95.
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Counting Problems
4. (This is problem 1 on the 1966 IMO) In a math contest, three problems, A, B,
and C were posed. Among the participants there were 25 who solved at least
one problem. Of all the participants who did not solve problem A, the number
who solved problem B was twice the number who solved C. The number who
solved only problem A was one more than the number who solved A and at
least one other problem. Of all participants who solved just one problem, half
did not solve problem A. How many solved just problem B?
Solution: The answer is 6. Consider the Venn diagram below:
A
B
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x1
x3
x2
x5
x4
x6
x7
C
Letting x1 , . . . , x7 denote the number of students in each of the regions depicted, it follows that
x1 +x2 +x3 +x4 +x5 +x6 +x7 = 25
+x3
−x6 −2x7 = 0
x1 −x2
−x4 −x5
= 1
x1
−x3
−x7 = 0
Subtracting equation 4 from the sum of the first three yields
x1 + 3x3 = 26.
Only the pairs (x1 , x3 ) = (2, 8), (5, 7), (8, 6), (11, 5), (14, 5), (17, 3), (20, 2),
(23, 1), (26, 0). However, since 1 + x2 + x4 + x5 = x3 + x7 = x1 , it follows that
x1 < 11. On the other hand, x1 > x3 . This means that only the pair (8, 6)
can work. Indeed, x1 = 8, x3 = 6, x7 = 2, x6 = 2, and x2 + x4 + x5 = 7.
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Counting Problems
5. A falling number is an integer whose decimal representation has the property
that each digit except the units digit is larger than the one to its right. For
example 96521 is a falling number but 89642 is not. How many n-digit falling
numbers are there, for n = 1, 2, 3, 4, 5, 6, and 7?
Solution: Each collection of distinct digits can be arranged to form a falling
number is
one way.
hence the number of such numbers with 2, 3, . . .
exactly
10
10
10
,
,
,
, ... for a total of 210 − 1 = 1023.
digits is 10
2
3
4
5
6. For how many integers n between 1 and 1000 does x2 + x − n factor into the
product of two linear factors with integer coefficients?
Solution: x2 + x − n = (x − a)(x + (a + 1)) implies a(a + 1) = n, and there
31 such cases less than 1000.
7. Twelve lattice points are arranged along the edges of a 3 × 3 square as shown.
How many triangles have all three of their vertices among these points?
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Solution: Note that
three
each
element subset except the co-linear ones de4
termine a triangle: 12
−
4
= 204.
3
3
8. How many numbers can be expressed as a sum of four distinct members of the
set {17, 21, 25, 29, 33, 37, 41}?
Solution: Each member of {17, 21, 25, 29, 33, 37, 41} is one more than a multiple of four. Therefore, any sum of four of them is a multiple of 4. The smallest
such number is 17 + 21 + 25 + 29 = 92 and the largest is 29 + 33 + 37 + 41 = 140
+1 =
and all the multiples of 4 between them are obtainable. There are 140−92
4
48
+
1
=
13
such
numbers.
4
Alternatively, we transform the problem into a simpler one. Because 17 =
4 · 4 + 1, 21 = 4 · 5 + 1, . . . , 41 = 4 · 10 + 1, it makes sense to set up a correspondence between R and the set of numbers generated by {4, 5, 6, 7, 8, 9, 10}
OR between R and those numbers generated by {−3, −2, −1, 0, 1, 2, 3}. The
set R in this case is just {−6, −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5, 6}, which has
13 members.
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Counting Problems
9. How many numbers can be obtained as the product of two or more of the
numbers 3, 4, 4, 5, 5, 6, 7, 7, 7?
Solution: Factor the product n of members of G into primes to get something
of the form n = 2i 3j 5k 7l . The exponent i is odd if and only if the 6 appears
in the product. The number of 5’s and 7’s in Sn is just k and l respectively
and the number of 4’s is b 2i c. The number of 6’s is i − 2b 2i c, and the number
of 3’s is j minus the number of 6’s. Thus the number of members of R is the
number alternative ways to treat the various values. We can include the 3 or
not, include the 6 or not, include 0, 1, or 2 of the 4’s, 0,1, or 2 of the 5’s, and
0,1,2, or 3 of the 7’s. This number is
2 · 2 · 3 · 3 · 4 − 1 − 5 = 138.
10. How many of the first 80 positive integers are expressible as a sum of three or
fewer members of the set {30 , 31 , 32 , 33 , 34 } if we are allowed to use the same
power more than once. For example, 5 can be represented, but 8 cannot.
Solution: The number of powers of 3 used is just the sum of the ternary
digits. It is useful therefore to consider the numbers from 1 to 26, 27 to 53,
and 54 to 80. Numbers in the range 1 to 26 have ternary representation of
the form (a2 a1 a0 )3 . How many of these satisfy a2 + a1 + a0 ≤ 3? There are
16 such numbers. Those in the range 27 to 53 all have the form (1a2 a1 a0 )3 .
There are 10 for which a2 + a1 + a0 ≤ 2. The number in the range 54 to 80
have the form (2a2 a1 a0 )3 . Only 4 of these satisfy a2 + a1 + a0 ≤ 1. Hence there
are 16 + 10 + 4 = 30 such numbers altogether.
11. (1994 UNC Charlotte Comprehensive Exam) How many integers can be expressed as a sum of two or more different members of the set {0, 1, 2, 4, 8, 16, 31}?
Solution: The 0 in the set means that we should count the individual elements
among the sums. If the 31 was a 32, we’d have unique (binary) representation.
Hence the number of sums is just 26 minus 1(the empty sum) minus the duplicated values, of which there is just one. Hence there are 64 − 2 = 62 members
of R.
12. Each of the numbers 1, 2, 3, 4, 5, . . . , 62 is achievable, and 62 is certainly the
largest member of R.
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Counting Problems
13. John has 2 pennies, 3 nickels, 2 dimes, 3 quarters, and 8 dollars. For how
many different amounts can John make an exact purchase?
Solution: We’ll count achievable amounts less than $1, and multiply by 9,
then add in the 9 values 9.00, 9.01, 9.02, 9.05, 9.06, 9.07, 9.10, 9.11, 9.12. They
are {0, 1, 2, 5, 6, 7, 10, 11, 12, . . .}, exactly three of every five consecutive values.
So, counting 0, there are 53 (100) = 60 such values. Hence there are 9×60−1 =
539, since we don’t count the value 0. Now adding in the nine uncounted
values, we get 539 + 9 = 548.
14. What is the size of the largest subset, S, of {1, 2, 3, . . . , 50} such that no pair
of distinct elements of S has a sum divisible by 7?
Solution: S cannot have both an a ≡ 1(mod7) and a b ≡ 6(mod7). therefore,
put the members of S into one of four boxes b0 , b1 , b2 , b3 according to number
a goes into box i if a is congruent to i or 7 − i. Now box b0 has at most 1 b1
has at most 8 (1,8,15,22,29,36,43,50) and the other two boxes have at most 7
for a total of 8 + 7 + 7 + 1 = 23.
15. How many ordered pairs (m, n) of positive integers are solutions to
4
m
+ n2 = 1?
Solution: Simplify to getmn − 2m − 4n = 0 or mn − 2m − 4n + 8 = 8 and
factor to get n-2=1 and m-4=8; n-2=2 and m-4=4; etc for four pairs.
16. How many paths consisting of a sequence of horizontal and/or vertical line
segments with each segment connecting a pair of adjacent letters in the diagram below, is the word CONTEST spelled out as the path is traversed from
beginning to end?
C
C
C O
C O N
O N T
C
C O C
C O N O C
O N T N O C
N T E T N O
T E S E T N
E S T S E T
C
O C
N O
C
Sol Imagine spelling the word backwards starting with the T in the middle
and working back to the C’s on the outside. Count the number of paths from
T to each letter in the array to get the array:
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Counting Problems
15
20 10
15 10 6
6 5
4
3
1 1 1
1
1
6
6
5
4
2
1
1
1
1
1
1
1
1
6
6 15
5 10 20
3 6 10 15
2 3
4
5 6
1 1
1
1 1 1
The sum of the numbers in the C positions is 64 + 64 − 1 = 127.
17. Recall that a Yahtzee Roll is a roll of five indistinguishable dice.
a. How many different Yahtzee Rolls are possible?
b. How many Yahtzee Rolls have exactly 3 different numbers showing?
18. How many six digit numbers
a. consist of six different digits?
b. consist of five different digits?
c. consist of three odd and three even digits?
d. have six different digits with no two even digits adjacent?
e. have four distinct odd digits and two distinct even digits which are not
adjacent?
19. How many four digit numbers abcd satisfy |a − d| = 2?
20. A box contains 2 pennies, 4 nickels, and 6 dimes. Six coins are drawn without
replacement, with each of the 12 coins having the same probability of being
chosen. What is the probability that the value of the coins is at least 50 cents?
21. Let S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.
a. How many five element subsets does the set have?
b. How many subsets of S have an odd number of members?
c. How many subsets of S have 1 as a member? d. How many subsets have 1
as a member and do not have 2 as a member?
22. Imagine that the 4 × 7 grid below represents the streets of a part of the city
where you live. You must walk 11 blocks to get from the lower left corner at
A to the upper right corner at B.
(a) How many different 11 block walks are there?
(b) How many 11 block walks avoid the terrible corner marked with the
bullet?
(c) How many 11 block walks go through the terrible corner?
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Counting Problems
(d) How many different 12 block walks are there from A to B?
(e) How many different 13 block walks are there from A to B?
B
•
A
23. How many positive integers less than 1000 have an odd number of positive
integer divisors?
24. How many integers can be obtained as a sum of two or more of the numbers
1, 3, 5, 10, 20, 50, 82?
Solution: (
Alan Tucker’s Applied Combinatorics, problem 40, page 179.) Each member
of G = {1, 3, 5, 10, 20, 50, 82} is larger than the sum of the smaller members
of G except 82. Therefore the only duplicated sums involve 82. The minimal
duplicated sums are 82 + 1 = 50 + 20 + 10 + 3 and 82 + 3 = 50 + 20 + 10 + 5.
The other duplicates is 88 and 86. Thus there are
!
7
2 −
− 1 − 4 = 116
1
7
different sums, since we don’t want to count the
sum, and the four duplicated values.
7
1
= 7 singletons, the empty
25. Sixteen lattice points are arranged in a 3 × 3 square as shown. How many
triangles have all three of their vertices among these points?
Solution:
16
3
− 10
4
3
− 4 = 516
7
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