Combinatorics
Math 336, Winter 2015
Professor Ben Richert
Exam 1
Solutions
Problem 1 (15pts) Quick computations and examples. You don’t need to simplify, or show much work
really, but do not leave binomials (or multinomials) in your answer.
(a–5pts) What is the coefficient of x2 y 4 z 2 in (3x − y − 2z)7 ?
Solution. Ha, this was a typo that should have said (x1 + x2 + x3 )8 , and
X
8
8
(x1 + x2 + x3 ) =
xn1 1 xn2 2 xn3 3
n
n
n
1
2
3
n +n +n =8
1
2
3
ni ∈N≥0
so
X
8
(3x + (−y) + (−2z)) =
n1 +n2 +n3 =8
ni ∈N≥0
hence the coefficient of x2 y 4 z 2 is
8
8!
36.
(3)2 (−1)4 (−2)2 =
2!4!2!
242
(b–5pts) What is the largest possible value of
Solution. The max is
10
b10/2c
=
8
(3x)n1 (−y)n2 (−2z)n3
n1 n2 n3
10
?
k
10
5
=
10!
.
5!5!
(c–5pts) Write out the first three terms an infinite sum which computes 91/3 . There should be no cube roots
in your answer.
Solution.
1/3
∞ X
1/3
1k 81/3−k
k
k=0
1/3 1/3
1/3 −2/3
1/3 −5/3
=
8 +
8
+
8
+ ···
0
1
2
9
= (1 + 8)
=2+
1/3
=
(1/3) 1 (1/3)(1/3 − 1) 1
+
+ ···
1! 4
2!
32
Problem 2 (15pts) Let S = {1, . . . , 5}. This problem considers chains and anti-chains of S.
(a–3pts) Give an example of a chain which contains 3 elements.
Solution.
n
o
∅, {2, 3}, {1, 2, 3, 5}
(b–3pts) Give an example of a maximal anti-chain which contains 3 elements, or explain why this cannot be
done.
n
o
Solution. {1}, {2}, {3, 4, 5}
(c–3pts) Given an example of an anti-chain containing 11 elements, or explain why this cannot be done.
Solution. By Sperner’s Theorem, the size of any antichains is bounded by
5
b5/2c
=
5
= 10,
2
so there is no antichain containing 11 elements.
(d–3pts) How many maximal chains of S are there?
Solution. There are 5! (recall that there is a bijection between the collection of maximal chains of
S and permutations of S).
(e–3pts) How many maximal chains are there which avoid the set {2, 3}?
Solution. There are 5! − 2 · 6 = 5! − 12. That is, there are two ways to build the chain up to {2, 3},
and 6 = 3! build towards {1, 2, 3, 4, 5} by adding 1, 4, and 5 afterwards, so we subtract this number
from the total number to count those which avoid {2, 3}.
Problem 3 (10pts) The lunch table in the math lounge is a long rectangle with 5 chairs on each side. The
north end of the table sits up against a whiteboard (the horror!) and there is no chair at the south end.
Suppose that 10 students show up to eat lunch. One of these, Eratosthenes brings a peanut butter and jelly
sandwich, but a second, Euler, is terribly allergic to peanuts and thus cannot sit directly across from or
besides Eratosthenes. How many ways can the students be arranged at the table? Explain.
Solution. There are 10! ways to seat the students without any restrictions (starting at the north-east seat,
there are 10 choices for the student in that seat, 9 choices for the student to sit in the seat directly south,
and so on). If Eratosthenes sits in a corner seat then there are 2 seats which Euler must avoid (that directly
across from Eratosthenes and the one beside him), so there are 8 choices for the seat directly across, 7 choices
for the neighboring seat, and then the remaining 7 students may take the remaining seats in any order, for
a total of 8 · 7 · 7! = 7 · 8! possible seatings. The same argument holds for each of the corners, so that there
are 4 · 7 · 8! possibilities. If Eratosthenes sits in a non-corner seat, then there are 3 seats which Euler must
avoid (that directly across from Eratosthenes and the seats to either side of him), so there are 8 choices for
the seat directly across, 7 choices for seat on his north side, 6 choices for the seat on his south side, and the
remaining 6 students may take the remaining seats in any order, for a total of 8 · 7 · 6 · 6! = 6 · 8!. There are
6 non-corner seats, for a total of 6 · 6 · 8! = 62 · 8! possibilities. So the number of seating arrangements such
that Euler won’t go into anaphylactic shock are 10! − 4 · 7 · 8! − 62 · 8!.
Problem 4 (20pts) The math club decides to take a field trip to the World Chili Cook-off, which features
20 different types of chili.
(a–10pts) Pooling their cash, the club decides to purchase 30 bowls of chili. How many different ways can
they do this? Explain.
Solution. Let xi be the number of bowls of chili of type i that the club buys.
Then weare
30 + 20 − 1
looking for the number of solutions to the equation x1 + · · · + x20 = 30, which is
=
30
49!
.
19!30!
(b–10pts) Suppose that of the 20 types of chili, 15 are designated anywhere between “mild” and “hot,” but
5 are designated as “nuclear.” Fearing the worst, the club decides that they can buy at most
2 bowls of chili designated “nuclear.” Now how many ways can they purchase their 30 bowls?
Explain.
Solution. If they math club decides to buy no nuclear rated chili, then there are only 15 available
styles and (as above) the number of possibilities is
30 + 15 − 1
44!
.
=
30
14!30!
If they decide to buy one nuclear bowl then there are 5 choices of the nuclear chili, and
29 + 15 − 1
43!
=
14!29!
29
choices of the other 29 bowls, so
5
43!
14!29!
2+5−1
6
possibilities. If they decide to risk two nuclear holocausts, then there are
=
= 15
2
2
choices for the nuclear chili (choose 2 bowls among 5 possibilities), and
28 + 15 − 1
42!
=
14!28!
28
choices of the remaining 28 bowls, so
15
42!
14!28!
possibilities. Thus there are
44!
43!
42!
+5
+ 15
14!30!
13!48!
14!28!
total possibilities.
2n
n
Problem 5 (15pts) Finish the following explanation. We want to show that
=2
+ n2 with a
2
2
2n
n
combinatorial argument, which comes down to showing that
and 2
+ n2 count the same thing. So
2
2
suppose
that the math club is made up of n pure mathematicians and n applied mathematicians. Then there
2n
are
different ways to choose 2 of the members to go pick up the pizza for the weekly club meeting.
2
But we could count the number of ways to choose two members to go pick up the pizza differently. If both
of the members we choose are applied mathematicians, then there
are . . .
n
Solution.
ways to choose a 2 people for the pizza run (since there are n applied people to choose from),
2
n
if both the members are to be pure mathematicians then there are similarly
possibilities. Finally, if
2
there is to be one pure and one applied mathematician to head out then there are n · n ways to proceed (n
choices of a pure mathematician and n choices
mathematician).
So the total number of ways
of
an applied
n
n
n
2
2
to choose 2 people to go pick-up the pizza is
+
+n =2
+ n as required.
2
2
2
Problem 6 (15pts) A certain professor decides to limit his coffee intake to at most 10 cups per week. As a
human being, he must drink at least 1 cup per day. Show that during the first 5 weeks of the quarter, there
is a string of consecutive days during which he drinks exactly 19 cups of coffee.
Solution. For i = 1, . . . , 35 let ai be the number of cups of coffee this addict has drunk by the end of the ith
day. Since he drinks at least 1 cup per day and at most 10 per weeks, we know that
1 ≤ a1 < a2 < · · · < a35 ≤ 50.
Now note that
20 ≤ a1 + 19 < · · · < a35 + 19 ≤ 69,
and hence the 70 integers
a1 , . . . , a35 , (a1 + 19), . . . , (a35 + 19)
are all elements of the set {1, . . . , 69}. By the Pigeon Hole Principle, there must be i, j ∈ {1, . . . , 35} and
k ∈ {1, . . . , 69} such that ai = aj + 19 = k (we note that ai 6= aj and ai + 19 6= aj + 19 for all i 6= j). If
i < j then ai < aj < aj + 19, a contradiction, so i > j. Thus ai − aj = 19, and this professor drank exactly
19 cups of coffee from the beginning of day i + 1 to the end of day j.