Download pdf - BinomialTheorem. Combinationswithrepetition. - Alexey Nikolaev · Pascal’sTriangle TheBinomial Theorem Combinationswith repetition Permutationswith repetition p.2 Permutationsandcombinations

Binomial Theorem.Combinations with repetition.

Pascal’s TriangleThe BinomialTheoremCombinations withrepetitionPermutations withrepetition

p. 2

Permutations and combinationsGiven a set with n elements.

The number of permutations of the elements the set:

P(n) = n!= n · (n− 1) · (n− 2) · . . . · 1

The number of r-permutations of the set:

P(n, r) =n!

(n− r)!= n · (n− 1) · (n− 2) · . . . · (n− r + 1)︸︷︷︸

product of r numbers

The number of unordered r-combinations (“n choose r”):�

nr

�

=P(n, r)P(r)

=n!

(n− r)! r!


p. 3

What are the properties of �nr

�?How does

�nr

�

change with r?

�

00

�

=0!

0! 0!= 1.

�

10

�

=1!

1! 0!= 1,

�

11

�

=1!

0! 1!= 1.

�

20

�

=2!

2! 0!= 1,

�

21

�

=2!

1! 1!= 2,

�

22

�

=2!

0! 2!= 1.

�

30

�

=3!

3! 0!= 1,

�

31

�

=3!

2! 1!= 3,

�

32

�

=3!

1! 2!= 3,

�

33

�

=3!

0! 3!= 1.


p. 4

Pascal’s Triangle

�

00

�

�

10

� �

11

�

�

20

� �

21

� �

22

�

�

30

� �

31

� �

32

� �

33

�

�

40

� �

41

� �

42

� �

43

� �

44

�

�

50

� �

51

� �

52

� �

53

� �

54

� �

55

�

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1


p. 5

Pascal’s TriangleThe numbers in Pascal’s Triangle are the coefficients of the polyno-mials of the form (x + y)n:

(x + y)0 = 1

(x + y)1 = 1x + 1y

(x + y)2 = 1x2 + 2x y + 1y2

(x + y)3 = 1x3 + 3x2 y + 3x y2 + 1y3

(x + y)4 = 1x4 + 4x3 y + 6x2 y2 + 4x y3 + 1y4

(x + y)n =�

n0

�

· xn +�

n1

�

· xn−1 y +�

n2

�

· xn−2 y2 + . . .+�

nn

�

· yn.

(x + y)n =n∑

k=0

�

nk

�

xn−k yk


p. 6


(x + y)0 = 1

(x + y)1 = 1x + 1y

(x + y)2 = 1x2 + 2x y + 1y2

(x + y)3 = 1x3 + 3x2 y + 3x y2 + 1y3

(x + y)4 = 1x4 + 4x3 y + 6x2 y2 + 4x y3 + 1y4

(x + y)n =�

n0

�

· xn +�

n1

�

· xn−1 y +�

n2

�

· xn−2 y2 + . . .+�

nn

�

· yn.

(x + y)n =n∑

k=0

�

nk

�

xn−k yk


p. 7


(x + y)0 = 1

(x + y)1 = 1x + 1y

(x + y)2 = 1x2 + 2x y + 1y2

(x + y)3 = 1x3 + 3x2 y + 3x y2 + 1y3

(x + y)4 = 1x4 + 4x3 y + 6x2 y2 + 4x y3 + 1y4

(x + y)n =�

n0

�

· xn +�

n1

�

· xn−1 y +�

n2

�

· xn−2 y2 + . . .+�

nn

�

· yn.

(x + y)n =n∑

k=0

�

nk

�

xn−k yk


p. 8


(x + y)0 = 1

(x + y)1 = 1x + 1y

(x + y)2 = 1x2 + 2x y + 1y2

(x + y)3 = 1x3 + 3x2 y + 3x y2 + 1y3

(x + y)4 = 1x4 + 4x3 y + 6x2 y2 + 4x y3 + 1y4

(x + y)n =�

n0

�

· xn +�

n1

�

· xn−1 y +�

n2

�

· xn−2 y2 + . . .+�

nn

�

· yn.

(x + y)n =n∑

k=0

�

nk

�

xn−k yk


p. 9

Coefficients of (x + y)n

Let’s prove

(x + y)n =�

n0

�

· xn +�

n1

�

· xn−1 y +�

n2

�

· xn−2 y2 + . . .+�

nn

�

· yn.

Example:

(x + y)3 =(x + y)(x + y)(x + y) =x x x+x x y + x y x + y x x+x y y + y x y + y y x+y y y

2n = 23 = 8 terms in total. The same as the number of the bitstrings of length 3.


p. 10


(x + y)n =(x + y) · (x + y) · . . . · (x + y)︸︷︷︸

n times

= ?

What is happening when we multiply (x + y) n times?

We get the sum of


p. 11


(x + y)n =(x + y) · (x + y) · . . . · (x + y)︸︷︷︸

n times

= ?

What is happening when we multiply (x + y) n times?

We get the sum of

x x x x x . . . x+yx x x x . . . x+xyx x x . . . x+yyx x x . . . x+x xyx x . . . x+. . .+y y y y y . . . y


p. 12


(x + y)n =(x + y) · (x + y) · . . . · (x + y)︸︷︷︸

n times

=

x · x · . . . · x︸︷︷︸

=xn

+

(x · . . . · x︸︷︷︸

=xn−1

) · y + . . .+ y · (x · . . . · x)+

(x · . . . · x︸︷︷︸

=xn−2

) · (y · y) + . . .+ (y · y) · (x · . . . · x︸︷︷︸

=xn−2

)+

. . .+y · y · . . . · y︸︷︷︸

=yn


p. 13


(x + y)n =(x + y) · (x + y) · . . . · (x + y)︸︷︷︸

n times

=

1 · xn +

n · xn−1 y +�

n2

�

· xn−2 y2 +

. . . +1 · yn


p. 14

The Binomial Theorem

(x + y)n =(x + y) · (x + y) · . . . · (x + y)︸︷︷︸

n times

=

�

n0

�

· xn +�

n1

�

· xn−1 y +�

n2

�

· xn−2 y2 + . . .+�

nn

�

· yn.

Shorter notation for the same thing:

(x + y)n =n∑

k=0

�

nk

�

xn−k yk.

This result is called The Binomial Theorem, and this is why the co-efficients,

�nk

�

, are also called the binomial coefficients.


p. 15


Let’s prove that�

n0

�

+�

n1

�

+ . . .+�

nn

�

= 2n.

(1+ 1)n =n∑

k=0

�

nk

�

1n−k1k =n∑

k=0

�

nk

�

· 1=n∑

k=0

�

nk

�

.

Therefore,n∑

k=0

�

nk

�

= 2n.


p. 16


Let’s prove that�

n0

�

+�

n1

�

+ . . .+�

nn

�

= 2n.

(1+ 1)n =n∑

k=0

�

nk

�

1n−k1k =n∑

k=0

�

nk

�

· 1=n∑

k=0

�

nk

�

.

Therefore,n∑

k=0

�

nk

�

= 2n.


p. 17

The Binomial Theorem�

n0

�

+�

n1

�

+ . . .+�

nn

�

= 2n.

Results like this are not very obvious.

Recall that�

nk

�

=n!

(n− k)! k!

So,n!

n! 0!+

n!(n− 1)! 1!

+n!

(n− 2)! 2!+ . . .+

n!0! n!

= 2n.

In this form, the result seems to be much harder to prove.

However, can you think of another way to prove this identity?(You can try to use double counting)


p. 18


Using the binomial theorem

(x + y)n =n∑

k=0

�

nk

�

xn−k yk,

prove that

n∑

k=0

2k�

nk

�

=

�

n0

�

+ 2�

n1

�

+ 22�

n2

�

+ . . .+ 2n�

nn

�

= 3n.


p. 19

Counting routes

You can go only North and East.Count the number of paths from (0,0) to (5, 3). Answer:

�5+33

�

.


p. 20

Counting routes

You can go only North and East.Count the number of paths from (0,0) to (5, 3). Answer:

�5+33

�

.


p. 21

Pascal’s Triangle Again

�

00

�

�

10

� �

11

�

�

20

� �

21

� �

22

�

�

30

� �

31

� �

32

� �

33

�

�

40

� �

41

� �

42

� �

43

� �

44

�

�

50

� �

51

� �

52

� �

53

� �

54

� �

55

�

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1


p. 22

Pascal’s Identity

�

n+ 1k

�

=�

nk− 1

�

+�

nk

�

Every number in Pascal’s triangle is equal to the sum of the twonumbers that are immediately above.


p. 23

Another Identity

Prove that for r ≤ n and r ≤ m:

�

m+ nr

�

=r∑

k=0

�

mr − k

��

nk

�


p. 24

Vandermonde’s IdentityProve that for r ≤ n and r ≤ m:

�

m+ nr

�

=r∑

k=0

�

mr − k

��

nk

�

We split the initial set of m+ n objsects into two arbitrary subsetsof m and n objects. After that, we can choose r objects from thetwo subsets in the following ways:

k subset of size m subset of size n

0 choose r choose none1 choose r − 1 choose 12 choose r − 2 choose 2. . .r choose 0 choose r

�

mr

��

n0

�

+�

mr − 1

��

n1

�

+

+�

mr − 2

��

n2

�

+ . . .+�

m0

��

nr

�

=r∑

k=0

�

mr − k

��

nk

�


p. 25

r-combinations without repetition

Recall that without repetitions, this is�n

r

�

.

For example, you have n books, but don’t have time to read all ofthem, and have to select only r books to read.

In how many ways can you do so?�

nr

�

=n!

(n− r)! r!

There are�n

r

�

ways to make the choice.


p. 26

r-combinations with repetition

There is a vending machine with 3 types of drinks, $1 each drink.

You have to spend $5.

| $ | $ | $ | $ | $ |


p. 27




| $ | $ | $ | $ | $ |


p. 28




$︸︷︷︸

1 drink

| $ $︸︷︷︸

2 drinks

| $ $︸︷︷︸

2 drinks


p. 29




| $ | $ | $ | $ | $ |


p. 30




$ $︸︷︷︸

2 drink

| $ $ $︸︷︷︸

3 drinks

|︸︷︷︸

none


p. 31


So, there are 5+ (3− 1) places that stand for 5 dollars and (3− 1)separators between the drinks’ types.

_ _ _ _ _ _ _

$ | $ $ | $ $

$ $ | $ $ $ |

$ $ $ $ $ | |�

72

�

=�

75

�

= 21 ways to buy 5 drinks


p. 32


So, there are 5+ (3− 1) places that stand for 5 dollars and (3− 1)separators between the drinks’ types.

_ _ _ _ _ _ _

$ | $ $ | $ $

$ $ | $ $ $ |

$ $ $ $ $ | |�

75

�

=�

72

�

= 21 ways to buy 5 drinks


p. 33


To select r objects out of n with repetitions, there are�

r + n− 1r

�

=�

r + n− 1n− 1

�

ways.

_ _ _ _ _ _ _ _ _ _

* * * * * * | | | |

(r objects and n− 1 separator)

In other words, this is the number of r-combinations with repetitionfrom the set of n objects.


p. 34

r-permutations with repetition

We know that the number of r-permutations of n objects withoutrepetition is

n(n− 1)(n− 2) · . . . · (n− r + 1) =n!

(n− r)!

But, if repetitions are allowed, it is even easier, the simple productrule works just fine!

n · n · . . . · n= nr


p. 35

Summary

with repetitions?

r-combination No�n

r

�

r-combination Yes�r+n−1

r

�

=�r+n−1

n−1

�

r-permutation No P(n, r) =n!

(n− r)!

r-permutation Yes nr