The Proof (concluded)

The Multinomial Theorem

Theorem 14

(x₁ + x₂ + · · · + x_t)ⁿ

= 

0≤n1,n2,...,nt≤n n1+n2+···+nt=n

n!

n₁! n₂!· · · nt! xⁿ₁¹xⁿ₂² · · · xⁿ_t ^t.

• Expand (x₁ + x₂ + · · · + x_t)ⁿ.

• Each term in the expansion must have the form (coefficient) × xⁿ₁¹xⁿ₂² · · · xⁿ_t ^t,

where 0 ≤ n₁, n₂, . . . , n_t ≤ n and n₁ + n₂ + · · · + n_t = n.

The Proof (concluded)

• The coefficient of

xⁿ₁¹xⁿ₂² · · · xⁿ_t ^t

equals the number of ways to pick n₁ x₁’s, n₂ x₂’s, and so on.

• By formula (2) on p. 16, there are

 n

n1, n2, . . . , nt



=Δ n!

n1!n2!· · · nt!

ways.

Coefficient of a

b

c

d

in (a + 2b − 3c + 2d + 5)

• Make x1 = a, x₂ = 2b, x₃ = −3c, x4 = 2d, and x₅ = 5 symbolically.

• The coefficient of a²(2b)³(−3c)²(2d)⁵5⁴ is

 16

2, 3, 2, 5, 4



= 16!

2! 3! 2! 5! 4! = 302, 702, 400 by the multinomial theorem with n = 16.

• The desired coefficient is then

302, 702, 400 × 2³ × (−3)² × 2⁵ × 5⁴

= 435, 891, 456, 000, 000.

Distinct Objects into Identical Containers

Corollary 15 There are _(r!)^(rn)!nn! ways to distribute rn distinct objects into n identical containers so that each container contains exactly r objects.

• Consider (x₁ + x₂ + · · · + x_n)^rn.

– Let xi denote the containers (distinct, for now).

– Each object is associated with one x₁ + x₂ + · · · + x_n. – It means an object can be assigned to one of the n

containers.

• What does the coefficient of

x^r₁x^r₂ · · · x^r_n

Distinct Objects into Identical Containers (continued)

• It is the number of ways rn distinct objects can be distributed into n distinct containers, each of which contains r objects.

• By Theorem 14 (p. 72), it is

 rn

r, r, . . . , r

 Δ

= (rn)!

r! r! · · · r!.

• Finally, divide the above count by n! to remove the identities of the containers.

Distinct Objects into Identical Containers (concluded)

Corollary 16 _(r!)^(rn)!nn! is an integer.

• Immediate from Corollary 15 (p. 75).

An Alternative Proof of Corollary 16 (p. 77)

(rn)!

(r!)ⁿn!

= 1

n!

(rn)!

[r(n − 1) ]! r!

[r(n − 1) ]!

[r(n − 2) ]! r! · · · [r(1) ]!

[r(n − n) ]! r!

=

_n−1

k=0

_r(n−k)

r

 n!

=

n−1

k=0

_r(n−k)

r

 n − k =

n−1

k=0

[r(n − k) ]!

(n − k)r![ r(n − k − 1) ]!

=

n−1

k=0

r(n − k)[ r(n − k) − 1 ]!

(n − k)r[ r − 1 ]![ r(n − k − 1) ]! =

n−1

k=0

r(n − k) − 1 r − 1

 .

a B93902003) on September 20, 2004.

Distinct Objects into Identical Containers (continued)

• Take n = 3 and r = 2.

• So we have

(x₁ + x₂ + x₃)⁶ = (x⁶₁ + · · · + x⁶₃) +6

x⁵₁x₂ + · · · + x2x⁵₃ +15

x⁴₁x²₂ + · · · + x²₂x⁴₃ +20

x³₁x³₂ + · · · + x³₂x³₃ +30

x⁴₁x₂x₃ + · · · + x₁x₂x⁴₃ +60

x³₁x²₂x₃ + · · · + x1x²₂x³₃ +90x²₁x²₂x²₃.

An Example (concluded)

• Indeed, the coefficients are

6 6

 ,

 6 5, 1

 ,

 6 4, 2

 ,

 6 3, 3

 ,

 6 4, 1, 1

 ,

 6 3, 2, 1

 ,

 6 2, 2, 2

 , consistent with the multinomial theorem (p. 72).

• The coefficient of x²₁x²₂x³₃ is 90.

• Thus the desired count is 90

3! = 15.

Combinations (Selections) with Repetition

Theorem 17 Suppose there are n distinct objects and r ≥ 0 is an integer. The number of selections of r of these objects, with repetition, is

C(n + r − 1, r) =

n + r − 1 r

 .

• Note that the order of selection is not important.

• Imagine there are n distinct types of objects.

The Proof (continued)

• Permute

  r

xx · · · x

  n−1

| | · · · | .

• Think of the ith interval as containing the ith type of objects.

• So

xx | xxx | x | | | |

means, out of 7 distinct objects, we pick 2 type-1 objects, 3 type-2 objects, and 1 type-3 object.

The Proof (concluded)

• Our goal equals the number of permutations of

  r

xx · · · x

  n−1

| | · · · | .

• By formula (2) on p. 16, it is (r + n − 1)!

r! (n − 1)! =

n + r − 1 r



= C(n + r − 1, r).

Combinatorial Proof of the Hockeystick Identity (P. 39)^a Corollary 18 For m, n ≥ 0, _m

k=0

_n+k

k

 = _n+m+1

m

.

• The number of ways to select m objects out of n + 2 types is _n+m+1

m

 by Theorem 17 (p. 81).

• Alternatively, let us focus on how the objects of the first n + 1 types are chosen.

• There are _n+m

m

 ways to select m objects out of the first n + 1 types.

• There are _n+m−1

m−1

 ways to select m − 1 objects out of the first n + 1 types and 1 object out of the last type.

aContributed by Mr. Jerry Lin (B01902113) on March 13, 2014.

The Proof (concluded)

• There are _n+m−2

m−2

 ways to select m − 2 objects out of the first n + 1 types and 2 objects of the last type.

• . . ..

• So,

n + m m

 +

n + m − 1 m − 1

 +

n + m − 2 m − 2



+ · · · +

n + 0 0



=

n + m + 1 m

 .

Integer Solutions of a Linear Equation

The following three problems are equivalent:

1. The number of nonnegative integer solutions of x₁ + x₂ + · · · + xn = r.

2. The number of selections, with repetition, of size r from a collection of n distinct objects (Theorem 17 on p. 81).

3. The number of ways r identical objects can be distributed among n distinct containers.^a

They all equal _n+r−1

r

.^b

aThe case of distinct objects and identical containers will be covered on p. 277 (see p. 75 for a special case).

Application: The Multinomial Theorem (P. 72)

• The theorem is about the coefficient of xⁿ₁¹xⁿ₂² · · · xⁿ_t ^t in the expansion of

(x₁ + x₂ + · · · + x_t)^r.

• But how many distinct terms^a are there?

• Each term has the form xⁿ₁¹xⁿ₂² · · · xⁿ_t ^t such that – n₁ + n₂ + · · · + nt = r, and

– 0 ≤ n₁, n₂, . . . , n_t.

• For example, consider

r = 2.

Application: The Multinomial Theorem (continued)

• Now,

(x₁ + x₂ + x₃)² = x²₁ + x²₂ + x²₃ + 2x₁x₂ + 2x₁x₃ + 2x₂x₃. – E.g., the solution “n₁ = 1, n₂ = 1, n₃ = 0” to

n₁ + n₂ + n₃ = 2 contributes to the term x¹₁x¹₂x⁰₃ = x₁x₂.

– So there are 6 nonnegative integer solutions to n₁ + n₂ + n₃ = 2 because there are 6 terms.

Application: The Multinomial Theorem (concluded)

• The desired number of terms is therefore

r + t − 1 r

 . from the equivalencies on p. 86.

• Indeed, ₂₊₃₋₁

2

 = 6.

Positive Integer Solutions of a Linear Equation

• Consider

x₁ + x₂ + · · · + xn = r, where x_i > 0 for 1 ≤ i ≤ n.

• Define x^_i = x^Δ _i − 1.

• The original problem becomes

x^₁ + x^₂ + · · · + x^_n = r − n, where x^_i ≥ 0 for 1 ≤ i ≤ n

• The number of solutions is therefore (p. 86)

n + (r − n) − 1 r − n



=

r − 1 r − n



=

r − 1 n − 1



. (14)

Application: Subsets with Restrictions

How many n-element subsets of { 1, 2, . . . , r } contain no consecutive integers?

• Say r = 4 and n = 2.

• Then the valid 2-element subsets of { 1, 2, 3, 4 } are { 1, 3 }, { 1, 4 }, { 2, 4 }.

The Proof (continued)

• For each valid subset { i1, i₂, . . . , i_n }, where 1 ≤ i1 < i₂ < · · · < in ≤ r, define

d_k = i_k+1 − i_k.

• As “placeholders,” introduce

i₀ = 1, i_n+1 = r.

• Then, by telescoping,

d₀ + d₁ + · · · + dn = i_n+1 − i0 = r − 1.

The Proof (continued)

• Observe that

0 ≤ d0, d_n

2 ≤ d₁, d₂, . . . , d_n−1.

• Define

d^₀ =^Δ d₀,

d^_k =^Δ d_k − 2, k = 1, 2, . . . , n − 1, d^_n =^Δ d_n.

The Proof (concluded)

• So equivalently,

d^₀ + d^₁ + · · · + d^_n = r − 1 − 2(n − 1) with 0 ≤ d^₀, d^₁, . . . , d^_n.

• The answer to the desired number is (p. 86)

(n + 1) + (r − 1 − 2(n − 1)) − 1 r − 1 − 2(n − 1)



=

 r − n + 1 r − 2n + 1



=

r − n + 1 n



. (15)

Application: Political Majority

In how many ways can 2n + 1 seats in a parliament be divided among 3 parties so that the coalition of any 2 parties form a majority?

• If n = 2, there are 5 seats.

• Clearly, no party should have 3 or more seats.

• The only valid distribution of the 5 seats to 3 parties is:

2, 2, 1.

• The number of ways is therefore 3.

aRecall p. 68.

The Proof (continued)

• This is a problem of distributing identical objects (the seats) among distinct containers (the parties) (p. 86).

• So without the majority condition, the number is

3 + (2n + 1) − 1 2n + 1



=

2n + 3 2

 .

• Observe that the majority condition is violated if and only if a party gets n + 1 or more seats (why?).

The Proof (concluded)

• If a given party gets n + 1 or more seats, the number of ways of distributing the seats is

3 + n − 1 n



=

n + 2 2

 .

– Allocate n + 1 seats to that party before allocating the remaining n seats to the 3 parties.

– Then refer to p. 86 for the formula.

• The desired number of no dominating party is

2n + 3 2



− 3

n + 2 2



= n

2 (n + 1) =

n + 1 2



. (16)

Political Majority: An Alternative Proof

• Recall that the majority condition holds if and only if no party gets n + 1 or more seats.

• So each party can hold up to n seats.

• Give each party n slots to hold real seats.

• As there are 2n + 1 seats, there will be 3n − (2n + 1) = n − 1 empty slots in the end.

aContributed by Mr. Weicheng Lee (B01902065) on March 14, 2013.

Political Majority: An Alternative Proof (concluded)

• So the answer to the desired number is the number of ways to distribute the n − 1 empty slots to 3 parties.

• The count is (p. 86)

3 + (n − 1) − 1 n − 1



=

n + 1 n − 1



=

n + 1 2

 .

Integer Solutions of a Linear Inequality

• Consider

x₁ + x₂ + · · · + x_n ≤ r, where x_i ≥ 0 for 1 ≤ i ≤ n.

• It is equivalent to

x₁ + x₂ + · · · + x_n + x_n+1 = r, where x_i ≥ 0 for 1 ≤ i ≤ n + 1.

• The number of integer solutions of the original inequality is therefore (p. 86)

(n + 1) + r − 1 r



=

n + r r



. (17)

The Hockeystick Identity (P. 39) Reproved

• By Eq. (17) on p. 100, there are _n+1+m

m

 nonnegative integer solutions to

x₁ + x₂ + · · · + xn+1 ≤ m, m ≥ 0.

• By p. 86, there are _n+k

k

 nonnegative integer solutions to

x₁ + x₂ + · · · + x_n+1 = k.

• Any solution to x₁ + x₂ + · · · + x_n+1 ≤ m is a solution to x₁ + x₂ + · · · + x_n+1 = k for some 0 ≤ k ≤ m.

The Proof (concluded)

• The opposite is also true.

• It is also clear the correspondence is one-to-one.

• So m

k=0

n + k k



=

n + m + 1 m

 .

• This is exactly the hockeystick identity.

Compositions of Positive Integers

• Let m be a positive integer.

• A composition for m is a sum of positive integers whose order is relevant and which sum to m.

• For m = 3, the number of compositions is 4:

3, 2 + 1, 1 + 2, 1 + 1 + 1.

• For m = 4, the number of compositions is 8:

4, 3+1, 2+2, 1+3, 1+1+2, 1+2+1, 2+1+1, 1+1+1+1.

• Is the number of compositions for general m equal to

m−1

The Number of Compositions

Theorem 19 The number of compositions for m > 0 is 2^m−1.

• Every composition with i summands corresponds to a positive integer solution to

x₁ + x₂ + · · · + xi = m.

• So the number of solutions is _m−1

m−i

 by Eq. (14) on p. 90.

• The total number of compositions is therefore

m i=1

m − 1 m − i



= 2^m−1

An Alternative Proof for Theorem 19 (p. 104)

• Let f(m) denote the number of compositions for m > 0.

• A composition for m is either (1) m or (2) i plus a

composition for m − i (“i + · · · ”) for i = 1, 2, . . . , m − 1.

• Then

f (m) = 1 +

m−1

i=1

f (m − i) = 1 +

m−1

i=1

f (i).

• The above implies that f(m + 1) − f(m) = f(m) so f (m + 1) = 2f (m).

a B01902046) on March 7, 2013.

The Proof (concluded)

• As a result,

f (m) = 2^m−1f (1).

• Finally, as f(1) = 1 = 2⁰,

f (m) = 2^m−1.

A Third Proof for Theorem 19 (p. 104)

• Start with m x’s and m − 1 |’s.

• Consider this arrangement:

 2m−1 x| x | x | · · · | x .

• Think of the |’s as dividers.

• Now remove some of the |’s.

aContributed by Mr. Jerry Lin (B01902113) on March 6, 2014.

The Proof (concluded)

• For example,

xx| xxx | x | x means the composition

2 + 3 + 1 + 1 for 7.

• Each removal of some |’s leads to a unique composition.

• As there are

2^m−1

ways to remove the |’s, this is the number of

Palindromes of Positive Integers

• Let m be a positive integer.

• A palindrome for m is a composition for m that reads the same left to right as right to left.

– For m = 4, the number of palindromes is 4:

4 , 1 + 2 + 1, 2 + 2, 1 + 1 + 1 + 1.

– For m = 5, the number of palindromes is 4:

5 , 1 + 3 + 1, 2 + 1 + 2, 1 + 1 + 1 + 1 + 1.

– The center elements are boxed above.

Palindromes of Positive Integers (concluded)

• The numbers to the left of the center element mirror those to the right, and with the same sum.

• Palindrome is possibly the hardest form of wordplay.^a

• For example,^b

A man, a plan, a canal, Panama!

aBryson (2001, p. 228).

bIgnore the spaces and punctuation marks.

The Number of Palindromes

Theorem 20 The number of palindromes for m > 0 is 2^m/2.

• Assume m is even first.

• The central element of a composition of m can be m, m − 2, . . . , 2 or “+” (think of it as a 0).^a

• When the central element is m, the number of palindromes is clearly 1.

• Suppose the central element is some other even number 0 ≤ i < m.

aThe central element must be even (why?)!

The Proof (concluded)

• Then the numbers to its left sum to (m − i)/2.^a

• They form a composition (p. 103).

• Hence the number of palindromes is 2^{(m−i)/2−1} by Theorem 19 (p. 104).

• The total number of palindromes for m is thus 1 +

1 + 2 + 2² + · · · + 2^{(m−2)/2−1} + 2^m/2−1

= 2^m/2.

• Follow the same argument when m is odd to obtain a count of 2^(m−1)/2.

aBy symmetry, the numbers to its right automatically sum to (m −

Runs

• Consider a permutation of 10 Os and 5 Es:

O O E O O O O E E E O O O E O.

• It has 7 runs:

O O

run

E

run

O O O O  

run

E E E  

run

O O O  

run

E

run

O

run

.

• In general, a run is a maximal consecutive list of identical objects.

The Number of Runs

Theorem 21 There are

 m − 1 m − r/2

 n − 1 n − r/2

 +

 n − 1 n − r/2

 m − 1 m − r/2



ways that m identical objects of type 1 and n identical objects of type 2 can give rise to r runs.

• Suppose the run starts with a type-1 object.

• Let xi denote the number of type-1 objects in run i = 1, 3, . . . , 2r/2 − 1.

The Proof (continued)

• The number of runs with the said counts x₁, x₃, . . . equals the number of positive-integer solutions to

x₁ + x₃ + · · · + x_2r/2−1 = m.

– There are r/2 terms.

• There are

 m − 1

r/2 − 1



=

 m − 1 m − r/2



solutions by Eq. (14) on p. 90.

The Proof (continued)

• Now let xi denote the number of type-2 objects in run i = 2, 4, . . . , 2r/2.

• The number of runs with the said counts x₂, x₄, . . . equals that of positive-integer solutions to

x₂ + x₄ + · · · + x_2r/2 = n.

– There are r/2 terms.

• Similarly, the number of solutions equals

 n − 1

r/2 − 1



=

 n − 1 n − r/2

 .

The Proof (concluded)

• Therefore the number of runs that start with a type-1 object equals

 m − 1 m − r/2

 n − 1 n − r/2

 .

• Repeat the argument for the case where the 1st run starts with a type-2 object.

• The count is

 n − 1 n − r/2

 m − 1 m − r/2



(by swapping m and n).

The Catalan

Numbers (1838)

• A binomial random walk starts at the origin (p. 43).

• What is the number of ways it can end at the origin in 2n steps without being in the negative territory?

• A left move lowers the position, whereas a right move increases the position.

• So it is equivalent to the number of ways

  n

RR· · · R

  n

LL· · · L

can be permuted so that no prefix has more Ls than Rs.

aEug`ene Charles Catalan (1814–1894). It was known to Euler (1707–

The Catalan Numbers (concluded)

• For example,

  0

  1

  2

  1

  0

  1

  0

 1

R L R L R R L L .

Formula for the Catalan Number

The number is^b b_n =

2n n



−

 2n n − 1



= 1

n + 1

2n n



, n ≥ 1. (18) with b₀ = 1.

•

  n

RR · · · R

  n

LL· · · L can be permuted in _2n

n

 ways by formula (2) on p. 16.^c

• Some of the permutations are illegal, such as RLLLRR.

aAttributed to Jacques Touchard (1885–1968).

bThe subscript in b_n is n not 2n!

The Proof (continued)

• We now prove that  _2n

n−1

 of the permutations are illegal.

• For every illegal permutation, we consider the first L move that makes the particle land at −1.

– Such as RL L LRR.

• Swap L and R for this offending L and all earlier moves.

– Such as L R R LRR.

• The result is a permutation of

  n+1

RR· · · R

  n−1

LL· · · L .

The Proof (concluded)

• There are  _2n

n−1

 ways to permute

  n+1

RR· · · R

  n−1

LL· · · L by Eq. (2) on 16.

• But the correspondence is one-to-one between the permutations of

  n+1

RR· · · R

  n−1

LL· · · L

and illegal permutations (see next page).

• So there are  _2n 

The Reflection Principle

aAndr´e (1887).

A Simple Corollary

Corollary 22 For n ≥ 1,

b_n =

_n

i=0

_n

i

₂ n + 1 .

• See Eq. (13) on p. 62.

Application: No Return to Origin until End

What is the number of ways a binomial random walk that is never in the negative territory and returns to the origin the first time after 2n steps?

• Let n ≥ 1.

• The answer is bn−1.

Application: No Return to Origin until End (concluded)

What is the number of ways a binomial random walk returns to the origin the first time after 2n steps?

• Let n ≥ 1.

• The answer is

2b_n−1 = 1 2n − 1

2n n



. (19)

– It may return to the origin by way of the negative territory.

– It may return to the origin by way of the positive

Application: Nonnegative Partial Sums

What is the number of ways we can arrange n “+1” and n

“−1” such that all 2n partial sums are nonnegative?

• For example, the six partial sums of (1, 1, −1, 1, −1, −1) are (1, 2, 1, 2, 1, 0).

• Let n ≥ 1.

• The answer is bn by definition (p. 118).

• The number remains bn if we have only n − 1 “−1”.

– In the original problem, the last number must be −1.

– So it is “redundant.”

Application: Nonpositive Partial Sums

What is the number of ways we can arrange n “+1” and n

“−1” such that all 2n partial sums are nonpositive?

• For example, the six partial sums of (−1, −1, 1, −1, 1, 1) are (−1, −2, −1, −2, −1, 0).

• Let n ≥ 1.

• The answer is bn.

• The number remains bn if we have only n − 1 “+1”.

– In the original problem, the last number must be 1.

– So it is “redundant.”

Combinatorics and “Higher” Mathematics

For relaxation, General Bradley did algebra problems, and he worked at integral calculus when he was flying an airplane

— or flying in his airplane.

He said it relaxed him, made him think.

— Chet Hansen, Major, aide to 5-star General Omar Bradley (1893–1981)

Growth of Factorials

n n! n n!

0 1 8 40320

1 1 9 362880

2 2 10 3628800

3 6 11 39916800

4 24 12 479001600

5 120 13 6227020800 6 720 14 87178291200 7 5040 15 1307674368000

A Logplot (Base Two)

50 100 150 200

200 400 600 800 1000 1200

Logplot of n!

A Useful Lower Bound for n!

Lemma 23 n! > (n/e)ⁿ. Proof:

ln(n!) = ln 1 + ln 2 + ln 3 + · · · + ln n

=

n k=1

ln k

>

n k=1

 _k

k−1

ln x dx as ln x is increasing

=

 _n

0

ln x dx

= [ x ln x − x ]ⁿ_x=0

− n.

ln x

k k - 1

How Good Is the Bound?

250 500 750 1000125015001750 20

40 60 80 100

n! over lower bound

Good, but probably not of the same order as n!.

A Marginally Better Lower Bound

Lemma 24 n! > e(n/e)ⁿ. Proof:

ln(n!) = ln 1 + ln 2 + ln 3 + · · · + ln n

=

n k=2

ln k

>

n k=2

 _k

k−1

ln x dx

≥

 _n

1

ln x dx

= [ x ln x − x ]ⁿ_x=1

− n + 1.

A Useful Upper Bound for C(n, m)

Lemma 25 C(n, m) < (ne/m)^m for any 0 < m ≤ n.^a Proof:

C(n, m) = n!

(n − m)! m!

= n(n − 1) · · · (n − m + 1) m!

≤ n^m m!

< n^m

(m/e)^m by Lemma 23 (p. 133)

= (ne/m)^m.

Stirling’s Formula

(1730)

• The notation f(x) ∼ g(x) means

x→∞lim f (x)/g(x) = 1, i.e.,

f (x) = g(x) + o(g(x)) as x → ∞.^b

• Stirling’s formula says:

Theorem 26 n! ∼ √

2πn (n/e)ⁿ.

Corollary 27 e = limn→∞ n/(n!)^1/n.

aJames Stirling (1692–1770); but due to Abraham DeMoivre (1667–

1754)!

Goodness of Approximation to n!

200 400 600 800 1.0002

1.0004 1.0006 1.0008

n! over approximation

Approximation of C(n, m)

• Stirling’s formula can be used to approximate C(n, m) better than Lemma 25 (p. 137) under some conditions.

• For that purpose, a more refined Stirling’s formula is stated below without proof:^a

√2πn

n e

_n

e¹²ⁿ⁺¹¹ < n! < √

2πn

n e

_n

e¹²ⁿ¹ . (20)

aRobbins (1955).

The Proof (concluded)

• Now from bounds (20) on p. 140,

C(n, m)

= n!

(n − m)! m!

<

√2πn _n

e

_n

e¹²ⁿ¹

2π(n − m) 

n−me

_n−m

e^12(n−m)+11 √

2πm _m

e

_m

e^12m+11

= 1

√2π

 n m

_m  n n − m

_n−m n m(n − m)

×e1−12n−144(m−n)2−144mn (··· )(··· )(··· )

< 1

√2π

 n m

_m  n n − m

_n−m n

m(n − m) . (21)

Approximation of C(n, m), 1 ≤ m ≤ n/2

C(n, m)

> 1

√2π

 n m

_m  n n − m

_n−m n

m(n − m) e^{12n+1 −1 12(n−m) −1 12m1}

= 1

√2π

 n m

_m  n n − m

_n−m n

m(n − m) e12(n−m)(12n+1) −^−12m−1 1 12m

≥ 1

√2π

 n m

_m  n n − m

_n−m n

m(n − m) e12m(24m+1) −^−12m−1 1 12m

= 1

√2π

 n m

_m  n n − m

_n−m n

m(n − m) e^{− 16m+(24m+1)1}

> 1

√2π

 n m

_m  n n − m

_n−m n

m(n − m) e^{− 16m}. (22)

The Proof (continued)

• Combine inequalities (21) on p. 141 and (22) on p. 142 under 1 ≤ m ≤ n/2 to obtain

√1 2π

 n m

_m 

n n − m

_n−m 

n

m(n − m) e^−6m1

< C(n, m)

< 1

√2π

 n m

_m 

n n − m

_n−m 

n

m(n − m).

The Proof (concluded)

• So

C(n, m) ∼ 1

√2π

 n m

_m  n n − m

_n−m  n

m(n − m) (23)

as m → ∞ and n − m → ∞.

• An alternative formulation is C(n, m) ∼ 1

√2πpqn (p^pq^q)⁻ⁿ,

where p = m/n and q^Δ = 1^Δ − p.

Application: Probability of Return to Origin

• Suppose the binomial random walk has a probability of 2⁻¹ = 0.5 of going in either direction (p. 47).

– This is called a symmetric random walk.

• The number of ways it is at the origin after 2n steps is_2n

n

 by formula (4) on p. 44.^a

• The probability for this to happen is

_2n

n

 2²ⁿ ≈

√1

2π 2ⁿ 2ⁿ

2 n

2²ⁿ ≈

 1

πn = O

 1

√n



(24) by Eq. (23) on p. 144.

aWe have seen _2n

n

 many times before (e.g., p. 58, p. 62, p. 120, and

Application: Probability of Return to Origin (concluded)

• Suppose 100 U.S. Senators vote on a bill randomly.^a

• What is the probability of a tie?^b

• By Eq. (24) on p. 145, it equals

₁₀₀

50



2¹⁰⁰ = 0.0795892 ≈ 1 12.

• The probability is surprisingly high.

• It rises to 0.176197 with 20 Senators in late 18th century.

aDixit & Nalebuff (1993).

bWhich is broken by the Vice President.

Application: Deviation

• Consider the symmetric random walk again.

• Its average position at the end is 0.

• Assume n is even.

• Given c > 0, after n steps what is the probability for the walk to end at a position ≥ c√

n for n sufficiently large?

Application: Deviation (continued)

• The probability that the walk ends at position k after n

steps is 

n

n+k2

 2⁻ⁿ

by formula (4) on p. 44, where k is even.

• The probability that the position is at least c√ n is

n k=c√

n

 n

n+k2



2⁻ⁿ ≈ 1 2 −

c√

n

k=2

 n

n+k2



2⁻ⁿ

by Eq. (9) on p. 59.

– The integer k must be even.

Application: Deviation (concluded)

• But

1 2 −

c√

n

k=2

 n

n+k2



2⁻ⁿ ≥ 1

2 − c√ n 2

n

n2



2⁻ⁿ according to the unimodal property (p. 28).^a

– That k is even accounts for the 2 in the denominator.

• Finally, the desired probability is 1

2 − 2⁻ⁿc√ n 2

n

n2



≥ 1

2 − c

 1 2π by Eq. (24) on p. 145 for n sufficiently large.

aCorrected by Mr. Gong-Ching Lin (B00703082) on March 8, 2012

An Upper Bound for C(2n, n)

Lemma 28 _2n

n

 < 4ⁿ/√

nπ .

Proof: From inequality (21) on p. 141,

2n n



< 1

√2π

2n n

_n 

2n 2n − n

_2n−n 

2n n(2n − n)

= 1

√nπ 4ⁿ.

Note that Lemma 25 (p. 137) gives a much looser upper bound of (2e)ⁿ ∼ 5.43656ⁿ.

A Tight Bound for C(2n, n)

Lemma 29 _2n

n

 ∼ 4ⁿ/√

nπ .^a

• From inequality (22) on p. 142,

2n n



> 1

√2π

2n n

_n 

2n 2n − n

_2n−n 

2n

n(2n − n) e⁻⁶ⁿ¹

= 1

√nπ 4ⁿe⁻⁶ⁿ¹ .

• Finally, recall Lemma 28 (p. 150).

a −1/(8n) < √nπ _2n

/4ⁿ < 1 (Hipp & Mattner, 2008).

A Tight Bound for C(2n, n) (concluded)

First Return to Origin

What is the probability a symmetric binomial random walk returns to the origin the first time at step 2n?

• Formula (19) on p. 126 says the probability is 1

2n − 1

2n n



2⁻²ⁿ.

• The above probability is asymptotically

∼ 1

2√ n³π by Lemma 29 (p. 151).

a