EECS 203, Discrete Mathematics
Winter 2010, University of Michigan, Ann Arbor
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Your Frst name (print):
Circle your lecture section: 1 (TTH9)
2 (TTH12)
Circle your discussion section:
011
(Mandava M1:30)
012
(Wu F3:30)
013
(Wu W10:30)
014
(Chen M3:30) 015 (Mandava W1:30)
Solutions to Sample Exam for Exam 3
Date: April 20, 2010
Section 1.
Problem 1.
Which, if any, of the following statements are true? Circle the true answers on page 2.
(A) For any
x
∈
R
, and any integer
n
≥
0, (
e
x
+
e
−
x
)
n
=
∑
n
k
=0
(
n
k
)
e
(2
k
−
n
)
x
.
Ans:
True. Follows directly from the binomial theorem.
(B)
1
n
+1
(
2
n
n
)
=
1
2(2
n
+1)
(
2
n
+2
n
+1
)
when
n
≥
1.
Ans:
True. Follows from the de±nition of
(
n
k
)
=
n
!
k
!(
n
−
k
)!
.
(C)
∑
2
n
i
=0
(
2
n
i
)
= 4
n
.
Ans:
True. Follows from the binomial theorem.
(D)
(
n
k
)
≤
(
en
k
)
k
for all integers
n
and
k
with 0
≤
k
≤
n
.
Ans:
True. Proved in Lecture 23 (see the lecture note for details).
Problem 2.
Which, if any, of the following statements are true? Circle the true answers on page 2.
(A) The number of di²erent strings that can be made from permuting the characters in the word PROCHLORO-
COCCUS is 15!
/
(2!4!4!).
Ans:
True. The word has 15 occurrences of P, R, O, C, H, L, U, S. Those occurring more than once are
R
for 2 times,
O
4 times, and
C
4 times. The answer follows from the fact that on any permutation of the
15 occurrences of letters, permuting the di²erent occurrences of the same letter give the same word.
(B) Let
n
≥
0 and
m
≥
1 be integers. The number of di²erent ways to place
n
indistinguishable
balls in to
m
di²erent bins is
(
n
+
m
−
1
n
−
1
)
.
Ans:
False. Should be
(
n
+
m
−
1
m
−
1
)
.
(C) Let
n
≥
1. If 2
n
+ 1 pigeons are placed in
n
di²erent cages, then there must be a cage with at least 3
pigeons.
Ans:
True. Follows from P.H.P.
(D) The number of permutations on
{
1
,
2
,...,n
}
,
n
≥
3, so that 1 and 2 are in a
k
=
⌈
(
n
+1)
/
2
⌉
cycle is
n
!
k
(
k
+1)
.
Ans:
False. To determine a permutation of the desired property, we permute
{
3
,
4
,...,
}
, put 1 at the front,
and 2 in any place after 1 and before the
k
−
1’th number (not counting 1). The ±rst
k
elements in the
new sequence is the
k
-cycle. Thus the number of such permutations is (
n
−
2)!(
k
−
1)
> n
!
/
(
k
(
k
+ 1)).
1