k>0.
Triangulating a regular polygon with n sides
In how many different ways may we triangulate a regular polygon with n sides (an “n-gon”) with labeled vertices? (A triangulation divides the polygon into triangular regions via the addition of non-intersecting diagonals.)
Call this number Tn. The triangle has T3D 1 triangulation and the square has T4D 2 triangulations:
1 2
4 3
1 2
4 3 Also, the pentagon has T5D 5 triangulations:
5
Focus on any one side of the n-gon, say joining vertices 1 and 2. In any triangulation, that side will form one side of a triangle. Consider cases depending on the location of the third vertex of that triangle. There are six cases for an octagon:
1 2
4.1. Binomial and multinomial coefficients 149 If the third corner is labeled 8, then triangulate the remaining 7-sided figure (with corners at 2-3-4-5-6-7-8) in T7ways. If the third corner is labeled 7, then triangulate 1-7-8 in T3
ways and 2-3-4-5-6-7 in T6ways, for a total of T3T6triangulations. If the third corner is labeled 6, then triangulate 1-6-7-8 in T4ways and 2-3-4-5-6 in T5ways, for a total of T4T5
triangulations. Continuing this produces
T8D T7C T3T6C T4T5C T5T4C T6T3C T7: In general, the same idea gives
TnD Tn 1C T3Tn 2C T4Tn 3C C Tn 2T3C Tn 1: for n > 4. By defining T2 WD 1 we can write instead
TnD
n 1X
kD2
TkTn kC1 for n > 3, where T2WD 1. (4.3)
This is a nonlinear recurrence relation, but the techniques of Section 3.5 still work.
Let f .x/ D P
n>2Tnxn 2 be the OGF offTngn>2. That is, Tn is the coefficient of xn 2 in f .x/. (The discrepancy between the index n and the power n 2 makes the algebra come out somewhat cleaner.)
Multiply equation (4.3) by xn 2and sum over n > 3:
X
n>3
Tnxn 2DX
n>3 n 1X
kD2
TkTn kC1
!
xn 2: (4.4)
The left-hand side is f .x/ T2 D f .x/ 1. The right-hand side is a convolution that appears to be something close to Œf .x/2. Write out a few terms and see:
T2T2xC .T2T3C T3T2/x2C .T2T4C T3T3C T4T2/x3C D x
T2T2C .T2T3C T3T2/xC .T2T4C T3T3C T4T2/x2C D x
f .x/2
:
And so equation (4.4) becomes f .x/ 1D x f .x/2
or x
f .x/2
f .x/C 1 D 0:
Now (more magic with generating functions!) solve this equation for the unknown function f using the quadratic formula:
f .x/D . 1/˙p
. 1/2 4.x/.1/
2x D 1˙p
1 4x
2x :
Apply the extended binomial theorem top
1 4xto get
.1 4x/1=2DX
n>0
1=2 n
!
. 4/nxn:
But the sum for f .x/ is over n > 2, so Depending on which solution we take,
TnD ˙1
since (remember!) Tnis the coefficient of xn 2in f .x/. Exercise 14 asks you to show that the negative solution is the one we want, and also that it simplifies to
TnD 1
This section covered extensions of the binomial coefficients and the binomial theorem.
Both the multinomial coefficients and the multinomial theorem extend the binomial coef-ficients and binomial theorem, respectively, in a natural, combinatorial way. The extended binomial theorem represents an analytic extension of the binomial theorem. We used it to solve a nonlinear recurrence relation.
Exercises
1. The pro football season lasts 16 games. The list WWLTWWWWLWWWLLTW is the record of a team that won its first two games, lost its third, tied its fourth, etc., and finished with a record of 10-4-2 (10 wins, four losses, two ties).
(a) How many ways are there for a team to finish 10-4-2?
(b) How many ways in part (a) do not have consecutive losses?
(c) How many ways in part (a) have a longest winning streak of six games?
2. Consider the letters in the word DIVISIBILITY.
(a) How many different 12-lists can be formed by rearranging the letters?
(b) How many 12-lists in part (a) do not contain adjacent Is?
3. A university has 120 incoming freshman that still have to be assigned to on-campus housing. The only remaining dorm holds 105 students and contains 42 doubles (rooms housing two students) and seven triples (three students). In how many ways can the university select 105 students to house in this dorm and then arrange those students into roommate pairs and triples, without yet assigning them to rooms?
4. In the previous exercise, suppose the university gets approval to house temporarily the remaining 15 students among the dorm’s three lounges. Each lounge will house five students. How many ways are there for the university to assign all 120 students to rooms?
4.1. Binomial and multinomial coefficients 151 5. A mouse that lives in a hotel wants to travel from the ground floor entrance at location .0; 0; 0/to its nest on the 10th floor at location .12; 9; 10/. Each move the mouse makes is either one room north, one room east, or one floor up. For example, from .0; 0; 0/the mouse moves either to .1; 0; 0/ or .0; 1; 0/ or .0; 0; 1/, respectively. How many ways are there for the mouse to travel?
6. SupposejAj D 42. How many equivalence relations on A are there that have distinct equivalence classes of sizes 4, 7, 7, 8, 8, and 8?
7. Use the equivalence principle to prove the formula for the multinomial coefficients given in Theorem 4.1.1.
8. Give a combinatorial proof:
Xn
9. Give a combinatorial proof: k n 2
10. Give two proofs of the following, one combinatorial and one non-combinatorial: For n > 2, n.n 1/2n 2DX
12. Given a positive integer n, a composition of n is a list of positive integers that sum to n. For example, .3; 1; 1/ and .1; 3; 1/ and .1; 4/ and .5/ are each a composition of 5.
In general, how many compositions of n are possible?
13. Find the coefficient of xninp 1 8x.
14. Finish the demonstration of formula (4.6). Be sure to justify why the negative solution in equation (4.5) is the correct one.
15. The associative property of multiplication says that x.yz/D .xy/z. In other words, to compute the product xyz you could either find yz first then multiply that by x, or you could find xy first and then multiply that by z. Thus there are two ways to compute a product of three numbers via pairwise products and without changing the order of the numbers.
There are five ways to do this with a product of four numbers:
w.x.yz// w..xy/z/ .wx/.yz/ .w.xy//z ..wx/y/z
Let an equal the number of ways to do this with a product of n numbers. We just found that a3D 2 and a4D 5.
Derive a recurrence relation for anand then solve it to find a formula for an. 16. Define˚n
k
as the number of .nC k/-lists of the form .a1; a2; : : : ; anCk/where n of the elements are 1s and k of the elements are 1s and where for all i , the sum of the first i entries is nonnegative:
a1C a2C C ai >0 for all i2 Œn C k.
(a) Findn
by complete enumeration. Also, explain why˚n k
(c) Give a combinatorial proof:˚n k and k is this identity valid?
(d) Prove that˚n
The paper P´olya (1956), entitled “On picture-writing,” is a classic exposition by the master problem-solver George P´olya. In it he explains how generating functions can be easily derived from symbolic series (as we did at the beginning of Section 3.3) and also solves the problem of counting triangulations of the regular n-gon that we covered in this section.
Exercise 16 is from “Counting arrangements of 1’s and 1’s” by D. F. Bailey which appeared in Mathematics Magazine 69, April 1996, 128-131. His purpose was to provide a new derivation of the formulanC11 2nn
for the n-th Catalan number.
4.2 Fibonacci and Lucas numbers
The following recurrence relation defines the well-known Fibonacci numbers:
F0D 1 F1D 1
FnD Fn 1C Fn 2 for n > 2.
The first few Fibonacci numbers are shown below.
n 0 1 2 3 4 5 6 7 8 9 10 11 12 13
Fn 1 1 2 3 5 8 13 21 34 55 89 144 233 377
The same recurrence but with one change in the initial conditions defines the Lucas num-bers:
L0 D 2 L1 D 1
LnD Ln 1C Ln 2 for n > 2.
And the first few Lucas numbers are shown below.
n 0 1 2 3 4 5 6 7 8 9 10 11 12 13
Ln 2 1 3 4 7 11 18 29 47 76 123 199 322 521
Both the Fibonacci and Lucas numbers (though the Fibonacci more so) are quite celebrated in mathematics and elsewhere. This section mainly concentrates on the Fibonacci numbers but look to the exercises for results about the Lucas numbers.
4.2. Fibonacci and Lucas numbers 153