segment
Seattle, WA
Seattle, WA
Seattle, WA
Seattle, WA
Seattle, WA
Seattle, WA
Seattle, WA
Seattle, WA
Seattle, WA
Seattle, WA
Seattle, WA
Seattle, WA
Seattle, WA
Seattle, WA
Seattle, WA
Seattle, WA
Seattle, WA
Seattle, WA
Seattle, WA
Seattle, WA
Seattle
STR Number
22586 Y
23667 N
53135 N
53990 N
3748 N
63161 Y
23668 N
9938 Y
63234 N
31600 N
1219 N
9662 Y
36042 N
56818 N
57049 Y
36996 N
35200 N
33611 N
5317 N
14923 N
45656 N
10295 N
35201 N
27615 Y
22271 N
43792 N
17868 N
63948 Y
54959 Y
1221 N
6612 N
559
United Kingdom
Blue Fin Building
110 Southwark Street
London SE1 0TA
Phone: +44 (0)20 7922 1930
Fax: +44 (0)20 7922 1931
www.strglobal.com
United States
735 East Main Street
Hendersonville
TN 37075
Phone: +1 (615) 824 8664
Fax: +1 (615) 824 3848
www.str.c
United Kingdom
Blue Fin Building
110 Southwark Street
London SE1 0TA
Phone: +44 (0)20 7922 1930
Fax: +44 (0)20 7922 1931
www.strglobal.com
United States
735 East Main Street
Hendersonville
TN 37075
Phone: +1 (615) 824 8664
Fax: +1 (615) 824 3848
www.str.c
United Kingdom
Blue Fin Building
110 Southwark Street
London SE1 0TA
Phone: +44 (0)20 7922 1930
Fax: +44 (0)20 7922 1931
www.strglobal.com
United States
735 East Main Street
Hendersonville
TN 37075
Phone: +1 (615) 824 8664
Fax: +1 (615) 824 3848
www.str.c
How to Conduct a
Market Study
The STR SHARE Center
Supporting Hotel-related Academic Research and Education
2016 STR. All Rights Reserved.
Table of Contents
Typical Hotel and Tourism Industry Research
Projects
STR Ad-Hoc Industry Reports and Data Files
5.4 Generating Functions with Two or More Variables
139
f (x, y) = 3x 11 y 23 + 3x 11 y 22 + x 10 y 23 + 2x 11 y 21
+3x 10 y 22 + x 9 y 23 + 3x 11 y 20 + 2x 10 y 21 + 2x 8 y 23 + 2x 11 y 19 + x 10 y 20 + 3x 9 y 21 + x 8 y 22
+x 11 y 18 + 2x 10 y 19 + 2x 9
38
2 Basic Counting
Exercise 2.3.8 Show that n! 2n1 for all n.
Exercise 2.3.9 Find the number of ways of arranging the elements of [n] in such a
way that even and odd numbers must alternate.
Exercise 2.3.10 Suppose that n men and n women are to be seated
5.1 Review of Factoring and Partial Fractions
119
Choosing two distinct values of x will yield a system of two equations with two
unknowns, which we can then solve. We make the convenient choices of x = 1 and
x = 2.
x = 1 3 = 3A1 A1 = 1
x = 2 5 = 6A2 A2 =
3.5 Stars and Bars
79
A closely related concept is that of the Catalan numbers. The Catalan numbers
count the number of lattice paths from (0, 0) to (n, n) that do not cross the diagonal
(they are allowed to touch the diagonal). Using a similar technique
5.1 Review of Factoring and Partial Fractions
117
Alternately, if we can manipulate the standard factoring in algebra to achieve
the desired form. Namely,
f (x) = 3x 2 9x + 6 = 3(x 2 3x + 2) = 3(x 2)(x 1)
!
"
!
"
1
1
= 3( 2) 1 x (1)(1 x) = 6 1 x (1 x).
2
36
2 Basic Counting
Fig. 2.3 Variations on the
same table setting
Example 2.3.2 Find the number of visually distinct ways of seating n men and n
women around a circular table with 2n seats labeled 1, ., 2n in such a way that sexes
alternate.
Solution The
40
2 Basic Counting
to 1000 that are divisible by both 2 and 3. In other words, this is the set of integers
less than or equal to 1000 that are divisible by 6. Hence, Corollary 2.4.2 gives
"
!
1000
= 166.
|A B| =
6
By the Principle of Inclusion and Exclus
4.2 The Solution of Certain Distribution Problems
101
Since the Ai are disjoint sets, it follows that the number of distributions is given
by
#"
#
k " #"
!
k n2 1 n 1 + i 1
|Ai | =
.
i
i1
k1
i=1
i=1
k
!
!
Note that in the previous example, the roles of th
4.4 The Twelvefold Way
113
4.4 The Twelvefold Way
The problem of distributing balls into urns seems innocent. However, it actually provides a framework for a more general, abstract problem. This framework is described
as the Twelvefold Way. The Twelvefold
54
2 Basic Counting
(i)
(ii)
(iii)
(v)
Seat Alice at the small table. This breaks the cycle.
Choose one of the two remaining seats for Bob.
Place Chad at the last place of this table.
Choose and order seven of the remaining 12 guests to sit at the large t
3.2 Application: Hands in Poker
67
the value of the triple, the hand is called four of a kind. So the number of ways of
drawing a three of a kind can be computed as follows:
! "
(i) Choose the three values, there are 13
ways to do this;
3
!"
(ii) Choose o
3.1 Unordered Subsets of [n]
65
at ai . For each set of two pairs that overlap at ai , we remove one such pair. Thus,
we have removed 12 pairs from the 66 available. The remaining 54 pairs will be our
pigeons. Thus by the Pigeonhole Principle, there exist
46
2 Basic Counting
(i) Choose a row for the three Xs. Since these cannot be on a diagonal, there are
6 possibilities.
(ii) Choose an additional square for the fourth X. There are 6 ways to do this. Note
that this removes one possible row to place the thr
5.2 Review of Power Series
5.2
121
Review of Power Series
Series are studied in calculus, however, their full potential is often not appreciated
at that time. In combinatorics, we often take a different philosophy than that of
calculus. However, some of t
Chapter 3
The Binomial Coefficient
We now turn our attention to one of the most fundamental and useful notions in all
of combinatorics, the binomial coefficient. You may recall the binomial coefficient
from high-school algebra class. However, we will give
52
2 Basic Counting
(iv) Place seven of the remaining ten guests around the first table. There are
P (10, 7)/7 possibilities.
(v) Place three of the remaining three guests around the final table. There are
P (3, 3)/3 possibilities.
So, by the Multiplicati
50
2 Basic Counting
Theorem 2.7.2 Every permutation on [n] can be written as a product of disjoint
cycles.
There are two useful parameters that often appear in discussions of permutations
and their cycle decompositions. The first is the number of disjoint
5.2 Review of Power Series
123
First note that 0x 1 = 0. So, we can just as easily start the summation at n = 1.
Hence,
!
1
=
nx n1 .
2
(1 x)
n=1
Let m = n 1. When we change variables in a definite integral, we must also change
the bounds of the integral
3.7 Application: Cryptosystems and the Enigma
91
Solution The length of the message key can be at most the length of the message
itself. Thus even if the keyword is four letters, say MATH and n = 10, then the
message key would be MATHMATHMA. Hence there a
Chapter 6
Recurrence Relations
In this chapter, we examine the problem of determining concise formulas for the
recurrence relations found in earlier sections. You will find that the terminology and
techniques involved are quite similar to the terminology
6.1 Finding Recurrence Relations
149
Fig. 6.1 Sierpinski graphs
New recurrences can often be obtained from other recurrences. We now give an
example of this.
Example 6.1.3 Let cfw_Fn be the Fibonacci sequence, in other words, Fn = Fn1 +
Fn2 , where F0 =
3.5 Stars and Bars
77
(iv) Any such path must go from (3, 4) to (1, 2) and then from
! "(1, 2) to (7, 10).
The number of paths from (3, 4) to (1, 2) is given by 28 . The number of
! "
paths from (1, 2) to (7, 10) is given by 20
. Hence, by the Multiplicat
30
2 Basic Counting
Applying the inductive hypothesis yields
n
| n+1
i=1 Ai | = | i=1 Ai | + |An+1 |
=
n
!
i=1
|Ai | + |An+1 | =
n+1
!
i=1
|Ai |.
Alternatively, if x A
/ Aj for j = i. Therefore, x is counted once by
"i , then x
| ni=1 Ai | and once by ni