aarjaneiro/planetmath huggingface dataset:
generalized factorial

generalized factorial

Definition

.

The

[

x

,

d

]

n

is defined for any number

x

∈

ℂ

, any stepsize

d

∈

ℂ

and any

n

∈

ℤ

, except for

-

x

∈

{

d

,

2

⁢

d

,

…

,

n

⁢

d

}

, by

[

x

,

d

]

n

:=

{

∏

j

=

0

n

-

1

(

x

-

j

⁢

d

)

n

∈

ℕ

1

n

=

0

∏

j

=

1

-

n

1

x

+

j

⁢

d

-

n

∈

ℕ

,

-

x

∉

{

d

,

2

⁢

d

,

…

,

n

⁢

d

}

.

If

x

=

n

,

d

=

1

and

n

∈

ℕ

then

[

n

,

1

]

n

=

{

n

⁢

(

n

-

1

)

⁢

…

⁢

(

2

)

⁢

(

1

)

n

>

0

1

n

=

0

,

on which it follows that

[

n

,

1

]

n

=

n

!

. This is why the above definition generalizes the notion of the usual

factorial

.

Title

generalized factorial

Canonical

name

GeneralizedFactorial

Date of creation

2013-03-22 16:08:31

Last modified on

2013-03-22 16:08:31

Owner

gilbert_51126 (14238)

Last modified by

gilbert_51126 (14238)

Numerical id

7

Author

gilbert_51126 (14238)

Entry type

Definition

Classification

msc 05A10

Generated on Thu Feb 8 20:35:56 2018 by

LaTeXML
Hosoya’s triangle

Hosoya’s triangle

Hosoya’s

triangle

or the

Fibonacci triangle

is a triangular

arrangement

of numbers (like Pascal’s triangle) based on the

Fibonacci numbers

. Each number is the sum of the two numbers above in either the left diagonal or the right diagonal. The first few rows are:

1

1

1

2

1

2

3

2

2

3

5

3

4

3

5

8

5

6

6

5

8

13

8

10

9

10

8

13

21

13

16

15

15

16

13

21

⋮

⋮

⋮

(See

sequence

A058071 in Sloaen’s OEIS). The

recurrence relation

is

H

⁢

(

0

,

0

)

=

H

⁢

(

1

,

0

)

=

H

⁢

(

1

,

1

)

=

H

⁢

(

2

,

1

)

=

1

and

H

⁢

(

n

,

j

)

=

H

⁢

(

n

-

1

,

j

)

+

H

⁢

(

n

-

2

,

j

)

or

H

⁢

(

n

,

j

)

=

H

⁢

(

n

-

1

,

j

-

1

)

+

H

⁢

(

n

-

2

,

j

-

2

)

.

Thus, the two outermost diagonals are the Fibonacci numbers, while the numbers on the middle vertical line are the squares of the Fibonacci numbers. All the other numbers in the triangle are the

product

of two distinct Fibonacci numbers greater than 1. The row sums are the convolved Fibonacci numbers (A001629 in Sloane’s OEIS).

References

1

Haruo Hosoya, “Fibonacci Triangle”

The

Fibonacci Quarterly

14

2 (1976): 173 - 178

2

Thomas Koshy,

Fibonacci

and

Lucas Numbers

and Applications

. New York: Wiley & Sons (2001): 187 - 195

Title

Hosoya’s triangle

Canonical name

HosoyasTriangle

Date of creation

2013-03-22 18:07:47

Last modified on

2013-03-22 18:07:47

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

5

Author

PrimeFan (13766)

Entry type

Definition

Classification

msc 05A10

Synonym

Fibonacci triangle

Synonym

Hosoya triangle

Generated on Thu Feb 8 20:36:08 2018 by

LaTeXML
inductive proof of binomial theorem

inductive proof of binomial theorem

We prove the theorem for a ring. We do not assume a unit for the ring.
We do not need

commutativity

of the ring, but only that

a

and

b

commute.

When

n

=

1

, the result is clear.

For the inductive step, assume it holds for

m

. Then for

n

=

m

+

1

,

(

a

+

b

)

m

+

1

=

(

a

+

b

)

⁢

(

a

+

b

)

m

=

(

a

+

b

)

⁢

(

a

m

+

b

m

+

∑

k

=

1

m

-

1

(

m

k

)

⁢

a

m

-

k

⁢

b

k

)

⁢

by the inductive hypothesis

=

a

m

+

1

+

b

m

+

1

+

a

⁢

b

m

+

b

⁢

a

m

+

∑

k

=

1

m

-

1

(

m

k

)

⁢

a

m

-

k

+

1

⁢

b

k

+

∑

k

=

1

m

-

1

(

m

k

)

⁢

a

m

-

k

⁢

b

k

+

1

=

a

m

+

1

+

b

m

+

1

+

∑

k

=

1

m

(

m

k

)

⁢

a

m

-

k

+

1

⁢

b

k

+

∑

k

=

0

m

-

1

(

m

k

)

⁢

a

m

-

k

⁢

b

k

+

1

⁢

by combining terms

=

a

m

+

1

+

b

m

+

1

+

∑

k

=

1

m

(

m

k

)

⁢

a

m

-

k

+

1

⁢

b

k

+

∑

j

=

1

m

(

m

j

-

1

)

⁢

a

m

+

1

-

j

⁢

b

j

⁢

let j=k+1 in second sum

=

a

m

+

1

+

b

m

+

1

+

∑

k

=

1

m

[

(

m

k

)

+

(

m

k

-

1

)

]

⁢

a

m

+

1

-

k

⁢

b

k

⁢

by combining the sums

=

a

m

+

1

+

b

m

+

1

+

∑

k

=

1

m

(

m

+

1

k

)

⁢

a

m

+

1

-

k

⁢

b

k

⁢

from Pascal’s rule

as desired.

Title

inductive proof of binomial theorem

Canonical name

InductiveProofOfBinomialTheorem

Date of creation

2013-03-22 11:48:06

Last modified on

2013-03-22 11:48:06

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

21

Author

Mathprof (13753)

Entry type

Proof

Classification

msc 05A10

Generated on Thu Feb 8 20:36:11 2018 by

LaTeXML
Leibniz harmonic triangle

Leibniz harmonic triangle

The

Leibniz harmonic triangle

is a triangular

arrangement

of

fractions

in which the outermost diagonals consist of the

reciprocals

of the row numbers and each inner cell is the

absolute value

of the cell above minus the cell to the left. To put it algebraically,

L

⁢

(

r

,

1

)

=

1

n

(where

r

is the number of the row, starting from 1, and

c

is the column number, never more than

r

) and

L

⁢

(

r

,

c

)

=

L

⁢

(

r

-

1

,

c

-

1

)

-

L

⁢

(

r

,

c

-

1

)

.

The first eight rows are:

1

1

2

1

2

1

3

1

6

1

3

1

4

1

12

1

12

1

4

1

5

1

20

1

30

1

20

1

5

1

6

1

30

1

60

1

60

1

30

1

6

1

7

1

42

1

105

1

140

1

105

1

42

1

7

1

8

1

56

1

168

1

280

1

280

1

168

1

56

1

8

⋮

⋮

⋮

The denominators are listed in A003506 of Sloane’s OEIS, while the numerators, which are all 1s, are listed in A000012. The denominators of the second outermost diagonal are

oblong numbers

. The sum of the denominators in the

n

th row is

n

⁢

2

n

-

1

.

Just as Pascal’s triangle can be computed by using

binomial coefficients

, so can Leibniz’s:

L

⁢

(

r

,

c

)

=

1

c

⁢

(

r

c

)

.

This triangle can be used to obtain examples for the Erdős-Straus conjecture (

http://planetmath.org/ErdHosStrausConjecture

) when

4

|

n

.

References

1

A. Ayoub, “The Harmonic Triangle and the

Beta Function

”

Math. Magazine

60

4 (1987): 223 - 225

2

D. Darling, “Leibniz’ harmonic triangle” in

The Universal Book of Mathematics: From Abracadabra To Zeno’s paradoxes

. Hoboken, New Jersey: Wiley (2004)

Title

Leibniz harmonic triangle

Canonical name

LeibnizHarmonicTriangle

Date of creation

2013-03-22 16:47:21

Last modified on

2013-03-22 16:47:21

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

5

Author

PrimeFan (13766)

Entry type

Definition

Classification

msc 05A10

Synonym

Leibniz’ harmonic triangle

Synonym

Leibniz’s harmonic triangle

Synonym

Leibniz’z harmonic triangle

Generated on Thu Feb 8 20:36:14 2018 by

LaTeXML
multinomial theorem

multinomial theorem

A

multinomial

is a mathematical

expression

consisting of two or more terms, e.g.

a

1

⁢

x

1

+

a

2

⁢

x

2

+

…

+

a

k

⁢

x

k

.

The multinomial theorem provides the general form of the expansion of the powers of this
expression, in the process specifying the

multinomial coefficients

which are found in that expansion. The expansion is:

(

x

1

+

x

2

+

…

+

x

k

)

n

=

∑

n

!

n

1

!

⁢

n

2

!

⁢

⋯

⁢

n

k

!

⁢

x

1

n

1

⁢

x

2

n

2

⁢

⋯

⁢

x

k

n

k

(1)

where the sum is taken over all

multi-indices

(

n

1

,

…

⁢

n

k

)

∈

ℕ

k

that
sum to

n

.

The expression

n

!

n

1

!

⁢

n

2

!

⁢

⋯

⁢

n

k

!

occurring in the expansion is called

multinomial coefficient

and is denoted by

(

n

n

1

,

n

2

,

…

,

n

k

)

.

Title

multinomial theorem

Canonical

name

MultinomialTheorem

Date of creation

2013-03-22 13:13:05

Last modified on

2013-03-22 13:13:05

Owner

bshanks (153)

Last modified by

bshanks (153)

Numerical id

12

Author

bshanks (153)

Entry type

Theorem

Classification

msc 05A10

Related topic

BinomialFormula

Related topic

BinomialCoefficient

Related topic

GeneralizedLeibnizRule

Related topic

NthDerivativeOfADeterminant

Defines

multinomial

Defines

multinomial coefficient

Generated on Thu Feb 8 20:36:20 2018 by

LaTeXML
multinomial theorem (proof)

multinomial theorem (proof)

Proof.

The below proof of the

multinomial theorem

uses
the

binomial theorem

and

induction

on

k

.
In

addition

, we shall use

multi-index notation

.

First,
for

k

=

1

, both sides equal

x

1

n

. For the induction step,
suppose the multinomial theorem holds for

k

.
Then the binomial theorem and the induction

assumption

yield

(

x

1

+

⋯

+

x

k

+

x

k

+

1

)

n

=

∑

l

=

0

n

(

n

l

)

⁢

(

x

1

+

⋯

+

x

k

)

l

⁢

x

k

+

1

n

-

l

=

∑

l

=

0

n

(

n

l

)

⁢

l

!

⁢

∑

|

i

|

=

l

x

i

i

!

⁢

x

k

+

1

n

-

l

=

n

!

⁢

∑

l

=

0

n

∑

|

i

|

=

l

x

i

⁢

x

k

+

1

n

-

l

i

!

⁢

(

n

-

l

)

!

where

x

=

(

x

1

,

…

,

x

k

)

and

i

is a multi-index in

I

+

k

. To

complete

the proof, we need to show that
the sets

A

=

{

(

i

1

,

…

,

i

k

,

n

-

l

)

∈

I

+

k

+

1

∣

l

=

0

,

…

,

n

,

|

(

i

1

,

…

,

i

k

)

|

=

l

}

,

B

=

{

j

∈

I

+

k

+

1

∣

|

j

|

=

n

}

are equal.
The

inclusion

A

⊂

B

is clear since

|

(

i

1

,

…

,

i

k

,

n

-

l

)

|

=

l

+

n

-

l

=

n

.

For

B

⊂

A

, suppose

j

=

(

j

1

,

…

,

j

k

+

1

)

∈

I

+

k

+

1

,
and

|

j

|

=

n

. Let

l

=

|

(

j

1

,

…

,

j

k

)

|

. Then

l

=

n

-

j

k

+

1

,
so

j

k

+

1

=

n

-

l

for some

l

=

0

,

…

,

n

.
It follows that that

A

=

B

.

Let us define

y

=

(

x

1

,

⋯

,

x

k

+

1

)

and let

j

=

(

j

1

,

…

,

j

k

+

1

)

be a multi-index in

I

+

k

+

1

.
Then

(

x

1

+

⋯

+

x

k

+

1

)

n

=

n

!

⁢

∑

|

j

|

=

n

x

(

j

1

,

…

,

j

k

)

⁢

x

k

+

1

j

k

+

1

(

j

1

,

…

,

j

k

)

!

⁢

j

k

+

1

!

=

n

!

⁢

∑

|

j

|

=

n

y

j

j

!

.

This completes the proof.

□

Title

multinomial theorem (proof)

Canonical name

MultinomialTheoremproof

Date of creation

2013-03-22 13:41:55

Last modified on

2013-03-22 13:41:55

Owner

Koro (127)

Last modified by

Koro (127)

Numerical id

4

Author

Koro (127)

Entry type

Proof

Classification

msc 05A10

Generated on Thu Feb 8 20:36:24 2018 by

LaTeXML
Pascal’s rule proof

Pascal’s rule proof

We need to show

(

n

k

)

+

(

n

k

-

1

)

=

(

n

+

1

k

)

Let us begin by writing the left-hand side as

n

!

k

!

⁢

(

n

-

k

)

!

+

n

!

(

k

-

1

)

!

⁢

(

n

-

(

k

-

1

)

)

!

Getting a common

denominator

and simplifying, we have

n

!

k

!

⁢

(

n

-

k

)

!

+

n

!

(

k

-

1

)

!

⁢

(

n

-

k

+

1

)

!

=

(

n

-

k

+

1

)

⁢

n

!

(

n

-

k

+

1

)

⁢

k

!

⁢

(

n

-

k

)

!

+

k

⁢

n

!

k

⁢

(

k

-

1

)

!

⁢

(

n

-

k

+

1

)

!

=

(

n

-

k

+

1

)

⁢

n

!

+

k

⁢

n

!

k

!

⁢

(

n

-

k

+

1

)

!

=

(

n

+

1

)

⁢

n

!

k

!

⁢

(

(

n

+

1

)

-

k

)

!

=

(

n

+

1

)

!

k

!

⁢

(

(

n

+

1

)

-

k

)

!

=

(

n

+

1

k

)

Title

Pascal’s rule proof

Canonical name

PascalsRuleProof

Date of creation

2013-03-22 11:47:14

Last modified on

2013-03-22 11:47:14

Owner

akrowne (2)

Last modified by

akrowne (2)

Numerical id

10

Author

akrowne (2)

Entry type

Proof

Classification

msc 05A10

Classification

msc 81T13

Classification

msc 53C80

Classification

msc 82-00

Classification

msc 83-00

Classification

msc 81-00

Generated on Thu Feb 8 20:36:27 2018 by

LaTeXML
Pascal’s rule (bit string proof)

Pascal’s rule (bit string proof)

This proof is based on an alternate, but

equivalent

, definition of the

binomial coefficient

:

(

n

r

)

is the number of bit strings (

finite sequences

of

0

s and

1

s) of length

n

with exactly

r

ones.

We want to show that

(

n

r

)

=

(

n

-

1

r

-

1

)

+

(

n

-

1

r

)

To do so, we will show that both sides of the equation are counting the same set of bit strings.

The left-hand side counts the set of strings of

n

bits with

r

1

s. Suppose we take one of these strings and remove the first bit

b

. There are two cases: either

b

=

1

, or

b

=

0

.

If

b

=

1

, then the new string is

n

-

1

bits with

r

-

1

ones; there are

(

n

-

1

r

-

1

)

bit strings of this nature.

If

b

=

0

, then the new string is

n

-

1

bits with

r

ones, and there are

(

n

-

1

r

)

strings of this nature.

Therefore every string counted on the left is covered by one, but not both, of these two cases. If we add the two cases, we find that

(

n

r

)

=

(

n

-

1

r

-

1

)

+

(

n

-

1

r

)

Title

Pascal’s rule (bit string proof)

Canonical name

PascalsRulebitStringProof

Date of creation

2013-03-22 12:23:03

Last modified on

2013-03-22 12:23:03

Owner

vampyr (22)

Last modified by

vampyr (22)

Numerical id

5

Author

vampyr (22)

Entry type

Proof

Classification

msc 05A10

Generated on Thu Feb 8 20:36:29 2018 by

LaTeXML
proof of upper and lower bounds to binomial coefficient

proof of upper and lower bounds to binomial coefficient

Let

2

≤

k

≤

n

be

natural numbers

. We’ll first prove the

inequality

(

n

k

)

≤

(

n

⁢

e

k

)

k

.

We rewrite

(

n

k

)

as

(

n

k

)

=

n

⁢

(

n

-

1

)

⁢

⋯

⁢

(

n

-

k

+

1

)

k

!

=

(

1

-

1

n

)

⁢

⋯

⁢

(

1

-

k

-

1

n

)

⋅

n

k

k

!

Since each of the parenthesized factors lies between

0

and

1

, we have

(

n

k

)

<

n

k

k

!

Since all the terms of the series

e

k

=

∑

n

=

0

∞

k

n

/

n

!

are

positive

when

k

is a positive real number, each term must be smaller than the whole sum; in particular, this implies that, for any non-negative

integer

k

, we have

e

k

>

k

k

/

k

!

. Rearranging this slightly,

1

<

k

!

⁢

e

k

k

k

Multiplying this inequality by the previous inequality for the

binomial coefficient

yields

(

n

k

)

<

n

k

k

!

⋅

k

!

⁢

e

k

k

k

=

(

n

⁢

e

k

)

k

To conclude the proof we show that

∏

i

=

1

n

-

1

(

1

+

1

i

)

i

=

n

n

n

!

⁢

∀

n

≥

2

∈

ℕ

.

(1)

∏

i

=

1

n

-

1

(

1

+

1

i

)

i

=

∏

i

=

1

n

-

1

(

i

+

1

)

i

i

i

=

∏

i

=

2

n

i

i

-

1

(

∏

i

=

1

n

-

1

i

i

-

1

)

⋅

(

n

-

1

)

!

Since each left-hand factor in (

1

) is

<

e

, we have

n

n

n

!

<

e

n

-

1

.
Since

n

-

i

<

n

⁢

∀

1

≤

i

≤

k

-

1

, we immediately get

(

n

k

)

=

∏

i

=

2

k

-

1

(

1

-

1

i

)

k

!

<

n

k

k

!

.

And from

k

≤

n

⇔

(

n

-

i

)

⋅

k

≥

(

k

-

i

)

⋅

n

⁢

∀

1

≤

i

≤

k

-

1

we obtain

(

n

k

)

=

n

k

⋅

∏

i

=

1

k

-

1

n

-

i

k

-

i

≥

(

n

k

)

k

.

Title

proof of upper and lower bounds to binomial coefficient

Canonical name

ProofOfUpperAndLowerBoundsToBinomialCoefficient

Date of creation

2013-03-22 13:49:06

Last modified on

2013-03-22 13:49:06

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

12

Author

rspuzio (6075)

Entry type

Proof

Classification

msc 05A10

Generated on Thu Feb 8 20:36:36 2018 by

LaTeXML
Singmaster’s conjecture

Singmaster’s conjecture

Conjecture. (David Singmaster). With the exception of the number 1, no

positive integer

appears in Pascal’s triangle more than twelve times.

Numbering the top row of Pascal’s triangle (the tip with the single instance of 1) as row 0, and leftmost column as row 0, it is clear that each

integer

n

>

2

occurs at least twice, specifically, at positions

(

n

,

1

)

and

(

n

,

n

-

1

)

. Singmaster was able to figure out that when there is a solution to

n

=

(

F

2

⁢

k

⁢

F

2

⁢

k

+

1

F

2

⁢

k

-

1

⁢

F

2

⁢

k

-

1

)

-

1

=

(

F

2

⁢

k

⁢

F

2

⁢

k

+

1

-

1

F

2

⁢

k

-

1

⁢

F

2

⁢

k

)

(with

F

i

being the

i

th

Fibonacci number

) the number

n

occurs six times in Pascal’s triangle, and he showed that there are infinitely many such numbers. Empirical evidence suggests the actual maximum of instances for a number to occur in Pascal’s triangle may be less than twelve: Pascal’s triangle has been computed to millions of rows and no number (besides 1) has been encountered more than eight times.

Title

Singmaster’s conjecture

Canonical name

SingmastersConjecture

Date of creation

2013-03-22 18:51:10

Last modified on

2013-03-22 18:51:10

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

4

Author

PrimeFan (13766)

Entry type

Conjecture

Classification

msc 05A10

Generated on Thu Feb 8 20:36:38 2018 by

LaTeXML
some formulas involving rising factorial

some formulas involving rising factorial

Recall that, for

a

∈

ℂ

and

n

a

nonnegative integer

, the

rising factorial

(

a

)

n

is defined by

(

a

)

n

=

∏

k

=

0

n

-

1

(

a

+

k

)

.

The following results hold regarding the rising factorial:

•

For all

a

∈

ℂ

, we have

(

a

)

0

=

1

.

•

For all nonnegative integers

n

, we have

(

1

)

n

=

n

!

.

•

The

binomial coefficients

are given by

(

a

n

)

=

(

-

1

)

n

⁢

(

-

a

)

n

n

!

.

•

The rising factorial relates to the

gamma function

. One

relation

is given by the formula

(

a

)

n

=

Γ

⁢

(

a

+

n

)

Γ

⁢

(

a

)

.

This formula can be used to extend the definition of rising factorial so that

n

can be any

complex number

provided that

a

+

n

is not a

nonpositive integer

.

•

Another relation between the rising factorial and the gamma function is given by

Γ

⁢

(

a

)

=

lim

n

→

∞

⁡

n

!

⁢

n

a

-

1

(

a

)

n

.

Title

some formulas involving rising factorial

Canonical name

SomeFormulasInvolvingRisingFactorial

Date of creation

2013-03-22 17:49:12

Last modified on

2013-03-22 17:49:12

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

4

Author

Wkbj79 (1863)

Entry type

Result

Classification

msc 05A10

Generated on Thu Feb 8 20:36:41 2018 by

LaTeXML
sum of powers of binomial coefficients

sum of powers of binomial coefficients

Some results exist on

sums of powers of binomial coefficients

. Define

A

s

as follows:

A

s

⁢

(

n

)

=

∑

i

=

0

n

(

n

i

)

s

for

s

a

positive integer

and

n

a

nonnegative integer

.

For

s

=

1

, the

binomial theorem

implies that the sum

A

1

⁢

(

n

)

is simply

2

n

.

For

s

=

2

, the following result on the sum of the squares of the

binomial coefficients

(

n

i

)

holds:

A

2

⁢

(

n

)

=

∑

i

=

0

n

(

n

i

)

2

=

(

2

⁢

n

n

)

that is,

A

2

⁢

(

n

)

is the

n

th

central binomial coefficient

.

Proof:

This result follows immediately from the

Vandermonde identity

:

(

p

+

q

k

)

=

∑

i

=

0

k

(

p

i

)

⁢

(

q

k

-

i

)

upon choosing

p

=

q

=

k

=

n

and observing that

(

n

n

-

i

)

=

(

n

i

)

.

Expressions for

A

s

⁢

(

n

)

for larger values of

s

exist in terms of

hypergeometric functions

.

Title

sum of powers of binomial coefficients

Canonical name

SumOfPowersOfBinomialCoefficients

Date of creation

2013-03-22 14:25:43

Last modified on

2013-03-22 14:25:43

Owner

Andrea Ambrosio (7332)

Last modified by

Andrea Ambrosio (7332)

Numerical id

7

Author

Andrea Ambrosio (7332)

Entry type

Result

Classification

msc 05A10

Classification

msc 11B65

Generated on Thu Feb 8 20:36:43 2018 by

LaTeXML
upper and lower bounds to binomial coefficient

upper and lower bounds to binomial coefficient

Given two

integers

n

,

k

>

0

such that

k

≤

n

, we have the following

inequalities

for the

binomial coefficient

(

n

k

)

:

(

n

k

)

≤

n

k

k

!

(

n

k

)

≤

(

n

⋅

e

k

)

k

(

n

k

)

≥

(

n

k

)

k

Here

e

is the base of

natural logarithms

.
Also, for large

n

,

(

n

k

)

≈

n

k

k

!

.

Title

upper and lower bounds to binomial coefficient

Canonical name

UpperAndLowerBoundsToBinomialCoefficient

Date of creation

2013-03-22 13:29:53

Last modified on

2013-03-22 13:29:53

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

6

Author

rspuzio (6075)

Entry type

Theorem

Classification

msc 05A10

Generated on Thu Feb 8 20:36:48 2018 by

LaTeXML
crazy dice

crazy dice

It is a standard exercise in elementary

combinatorics

to the number of ways of rolling any given value with 2 fair 6-sided dice (by taking the of the two rolls). The below table the number of such ways of rolling a given value

n

:

n

# of ways

2

1

3

2

4

3

5

4

6

5

7

6

8

5

9

4

10

3

11

2

12

1

A somewhat (un?)natural question is to ask whether or not there are any other ways of re-labeling the faces of the dice with

positive integers

that these sums with the same frequencies. The surprising answer to this question is that there does indeed exist such a re-labeling, via the

labeling

Die

⁢

1

=

{

1

,

2

,

2

,

3

,

3

,

4

}

Die

⁢

2

=

{

1

,

3

,

4

,

5

,

6

,

8

}

and a pair of dice with this labeling are called a set of

crazy dice

. It is straight-forward to verify that the various possible occur with the same frequencies as given by the above table.

Title

crazy dice

Canonical name

CrazyDice

Date of creation

2013-03-22 15:01:43

Last modified on

2013-03-22 15:01:43

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

7

Author

mathcam (2727)

Entry type

Definition

Classification

msc 05A15

Generated on Thu Feb 8 20:36:50 2018 by

LaTeXML
derangement

derangement

Let

S

n

be the

symmetric group

of order

n

≥

1

. A

permutation

ϕ

∈

S

n

without a

fixed point

(that is,

ϕ

⁢

(

i

)

≠

i

for any

i

∈

{

1

,

…

,

n

}

) is called a

derangement

.

In combinatorial theory, one is usually interested in the number

d

⁢

(

n

)

of derangements in

S

n

. It is clear that

d

⁢

(

1

)

=

0

,

d

⁢

(

2

)

=

1

, and

d

⁢

(

3

)

=

2

. It is also not difficult to calculate

d

⁢

(

n

)

for small

n

. For general

n

, one can appeal to the

principle of inclusion-exclusion

. Let

A

i

denote the

collection

of permutations that fix

i

. Then the collection of

A

of derangments in

S

n

is just the

complement

of

A

1

∪

A

2

∪

⋯

∪

A

n

in

S

n

. Let

C

=

{

A

1

,

…

,

A

n

}

and

I

k

be the

k

-fold

intersection

of members of

C

(that is, each member of

I

k

has the form

A

j

1

∩

⋯

∩

A

j

k

). We can interpret a member of

I

k

as a set of permutations that fix

k

elements

from

{

1

,

…

,

n

}

. The

cardinality

of each of these members is

(

n

-

k

)

!

. Furthermore, there are

(

n

k

)

members in

I

k

. Then,

d

⁢

(

n

)

=

|

A

|

=

|

S

n

|

-

|

A

1

∪

⋯

∪

A

k

∪

⋯

∪

A

n

|

=

n

!

-

[

∑

S

∈

I

1

|

S

|

-

⋯

+

(

-

1

)

k

⁢

∑

S

∈

I

k

|

S

|

+

⋯

+

(

-

1

)

n

⁢

∑

S

∈

I

n

|

S

|

]

=

n

!

-

[

(

n

1

)

⁢

(

n

-

1

)

!

-

⋯

+

(

-

1

)

k

⁢

(

n

k

)

⁢

(

n

-

k

)

!

+

⋯

+

(

-

1

)

n

⁢

(

n

n

)

⁢

(

n

-

n

)

!

]

=

n

!

-

[

n

!

1

!

-

⋯

+

(

-

1

)

k

⁢

n

!

k

!

+

⋯

+

(

-

1

)

n

⁢

n

!

n

!

]

=

n

!

⁢

∑

k

=

0

n

(

-

1

)

k

k

!

With this equation, one can easily derive the

recurrence relation

d

⁢

(

n

)

=

n

⁢

d

⁢

(

n

-

1

)

+

(

-

1

)

n

.

Apply this

formula

twice, we are led to another recurrence relation

d

⁢

(

n

)

=

(

n

-

1

)

⁢

[

d

⁢

(

n

-

1

)

+

d

⁢

(

n

-

2

)

]

.

Application

. A group of

n

men arrive at a party and check their hats. Upon departure, the hat-checker, being forgetful, randomly (meaning that the

distribution

of picking any hat out of all possible hats is a

uniform distribution

) hands back a hat to each man. What is the

probability

p

⁢

(

n

)

that no man receives his own hat? Does this probability go to

0

as

n

gets larger and larger?

Answer: According to the above calculation,

p

⁢

(

n

)

=

d

⁢

(

n

)

n

!

=

∑

k

=

0

n

(

-

1

)

k

k

!

.

Therefore,

lim

n

→

∞

⁡

p

⁢

(

n

)

=

1

e

≈

0.368

.

Title

derangement

Canonical name

Derangement

Date of creation

2013-03-22 16:05:04

Last modified on

2013-03-22 16:05:04

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

7

Author

CWoo (3771)

Entry type

Definition

Classification

msc 05A15

Classification

msc 05A05

Classification

msc 60C05

Generated on Thu Feb 8 20:37:02 2018 by

LaTeXML