MATHS SYMBOLS - OTHER OPERATIONS (2) - SUMMATION - Pi (PRODUCT) - FACTORIAL - SUM of FACTORIALS - SUBTRACTION of FACTORIALS - PRODUCT of FACTORIALS - DIVISION of FACTORIALS - POWER of FACTORIALS - PERMUTATIONS - POSSIBILITIES of COMBINATIONS - COMBINATIONS - BINOMIAL COEFFICIENT - BINOMIAL FORMULA - TETRATION - PENTATION - EXAMPLES and CALCULATIONS STEP by STEP

1. ENZO EXPOSYTO
MATHS
SYMBOLS
OTHER OPERATIONS - 2
Enzo Exposyto 1

2. n!
OTHER
OPERATIONS - 2
10
∑
k=1
k
3
∏
i=1
i
(
n
k)
4
2
3
3
3
Enzo Exposyto 2

3.
Enzo Exposyto 3

4. 1 - Summation 5
2 - Pi (Product) 13
3 - Factorial 21
4 - Binomial Coefficient 38
5 - Tetration 50
6 - Pentation 54
7 - SitoGraphy 59
Enzo Exposyto 4

5. SUMMATION
Enzo Exposyto 5

6. Summation - The Symbol
Enzo Exposyto 6
5 = the upper limit
of summation
this “i” represents
each addend
Σ = the Greek
letter sigma
i = the index
of summation
1 = the lower limit
of summation

7. Summation
∑ say: sigma
It represents summation
For example:
The expression means:
sum the values of i, starting at i = 1 and ending with i = 5.
This expression
has the same meaning.
Enzo Exposyto 7

8. Summation
∑ This expression
means:
sum the values of k,
starting at k = 5 and ending with k = 8.
This expression
has the same meaning.
Enzo Exposyto 8

9. Summation
∑ This expression
means:
sum the values of 2.k,
starting at k = 1 and ending with k = 3.
This expression
has the same meaning
Enzo Exposyto 9

10. Summation
∑ This expression (that is equivalent to 3 x 5)
means:
sum the value 5,
starting at 1 and ending with 3.
The expression (that is equivalent to 4.y)
has a similar meaning.
Enzo Exposyto 10

11. Summation
Data:
i xi
1 -1
2 2
3 1
4 3
Enzo Exposyto 11

12. Summation
Data:
i xi
1 -1
2 2
3 1
4 3
Enzo Exposyto 12

13. Pi
(PRODUCT)
Enzo Exposyto 13

14. Pi (Product) - The Symbol
Enzo Exposyto 14
4 = the upper limit
of product
1 = the lower limit
of product
i = the index
of product
Π = the Greek
letter pi
this “i” represents
each factor

15. Pi (Product)
∏ say: pi
It represents product
For example:
The expression means:
multiply the values of i, starting at i = 1 and ending with i = 4.
This expression, which is equivalent to 4! (see page 22),
has the same meaning.
Enzo Exposyto 15

16. Pi (Product)
∏ This expression
means:
multiply the values of i, starting at i = 2 and ending with i = 5.
This expression
has the same meaning.
Enzo Exposyto 16

17. Pi (Product)
∏ This expression
means:
multiply the values of 2.i,
starting at i = 1 and ending with i = 3.
This expression
has the same meaning
Enzo Exposyto 17

18. Pi (Product)
∏ This expression (that is equivalent to 53)
means:
multiply the value 5,
starting at 1 and ending with 3.
The expression (which is equivalent to z5)
has a similar meaning.
Enzo Exposyto 18

19. Pi (Product)
Data:
i xi
1 -1
2 2
3 1
4 3
Enzo Exposyto 19

20. Pi (Product)
Data:
i xi
1 -1
2 2
3 1
4 3
Enzo Exposyto 20

21. FACTORIAL
Enzo Exposyto 21

22. Factorial - 1
! factorial
For example:
2! = 2x1 = 2
3! = 3x2x1 = 6
4! = 4x3x2x1 = 24
5! = 5x4x3x2x1 = 120
Of course:
4! = 4x3x2x1 = 1x2x3x4
6! Say: 6 factorial
6 shriek
6 bang
6 crit
Enzo Exposyto 22

23. Factorial - 2
! If we define:
0! = 1
then,
with n integer ≥ 0,
we get:
(n+1)! = (n+1) x n!
or
n! = n x (n-1)!
For example:
5! = 5x4! = 5x4x3x2x1 = 120
4! = 4x3! = 4x3x2x1 = 24
3! = 3x2! = 3x2x1 = 6
2! = 2x1! = 2x1 = 2
1! = 1x0! = 1x1 = 1
Enzo Exposyto 23

24. Factorial - 3
! If there are n distinct elements into a set,
the factorial n! gives the number of ways in which
the n elements can be permuted:
the permutations of those elements.
For example:
Let A = {a, b, c}
Since A has 3 elements,
there are 3! = 6 possible permutations
of the elements of A:
{a, b, c}, {a, c, b}, (1st element: a)
{b, a, c}, {b, c, a}, (1st element: b)
{c, a, b}, {c, b, a} (1st element: c)
Enzo Exposyto 24

25. Factorial - 4
! A remarkable example:
Let A = {}
Since A has 0 elements,
there is 0! = 1 possible permutation
of the 0 elements of A:
{}
This is the consequence
of the definition 0! = 1
and of the convention for which
the product of no numbers is 1:
there is only one permutation
if the number of elements is zero.
Enzo Exposyto 25

26. Factorial - 5
! Addition:
a) n = k
n! + n! = 2 x n!
5! + 5! = 2 x 5!
0! + 0! = 2 x 0!
b) n > k
n! + k! = [nx(n-1)x(n-2)x … x1] + [kx(k-1)x(k-2)x … x1]
= [nx(n-1)x … x(k+1)]x(k!) + (k!)
= {[nx(n-1)x … x(k+1)]+1}x(k!)
5! + 3! = (5x4x3x2x1) + (3x2x1) = 5x4x(3x2x1) + (3x2x1)
= 5x4x(3!) + (3!) = [(5x4)+1] x 3!
Enzo Exposyto 26

27. Factorial - 6
! Subtraction:
a) n = k
n! - n! = 0
5! - 5! = 0
0! - 0! = 0
b) n > k
n! - k! = [nx(n-1)x(n-2)x … x1] - [kx(k-1)x(k-2)x … x1]
= [nx(n-1)x … x(k+1)]x(k!) - (k!)
= {[nx(n-1)x … x(k+1)]-1}x(k!)
5! - 3! = (5x4x3x2x1) - (3x2x1) = 5x4x(3x2x1) - (3x2x1)
= 5x4x(3!) - (3!) = [(5x4)-1] x 3!
Enzo Exposyto 27

28. Factorial - 7
! Multiplication:
a) n = k
n! * n! = (n!)2
5! * 5! = (5!)2
0! * 0! = (0!)2 = (1)2
b) n > k
n! * k! = [nx(n-1)x(n-2)x … x1] * [kx(k-1)x(k-2)x … x1]
= [nx(n-1)x … x(k+1)]x(k!) * (k!)
= [nx(n-1)x … x(k+1)]x(k!)2
5! * 3! = (5x4x3x2x1) * (3x2x1) = 5x4x(3x2x1) * (3x2x1)
= 5x4x(3!) * (3!) = (5x4) x (3!)2
Enzo Exposyto 28

29. Factorial - 8
! Division:
a) n = k
n! = 1
n!
5! = 1
5!
0! = 1
0!
Enzo Exposyto 29

30. Factorial - 9
! Division:
b1) n > k
n! = [nx(n-1)x(n-2)x … x1]
k! [kx(k-1)x(k-2)x … x1]
= [nx(n-1)x … x(k+1)]x(k!)
(k!)
= [nx(n-1)x … x(k+1)]
5! = 5x4x3x2x1 = 5x4x(3x2x1) = 5 x 4 x 3! = 5 x 4
3! 3x2x1 (3x2x1) 3!
Enzo Exposyto 30

31. Factorial - 10
! Division:
b2) n > k
k! = [kx(k-1)x(k-2)x … x1]
n! [nx(n-1)x(n-2)x … x1]
= k!
[nx(n-1)x … x(k+1)]x(k!)
= 1
[nx(n-1)x … x(k+1)]
= [nx(n-1)x … x(k+1)]-1
3! = 3x2x1 = (3x2x1) = 3! = 1 = [5x4]-1
5! 5x4x3x2x1 5x4x(3x2x1) 5x4x3! 5x4
Enzo Exposyto 31

32. Factorial - 11
! Power:
(n!)2 = n! x n!
= [nx(n-1)x(n-2)x … x1] x [nx(n-1)x(n-2)x … x1]
= (nxn)x(n-1)x(n-1)x(n-2)x(n-2)x … x1x1
= n2x(n-1)2x(n-2)2x … x12
(n!)k = n! x n! x … x n! (k times n!)
= …
= nkx(n-1)kx(n-2)kx … x1k
3!2 = 3! x 3! = (3x2x1) x (3x2x1) = (3x3)(2x2)(1x1) = 32 x 22 x12
Enzo Exposyto 32

33. Factorial - 12
nk k-possibilities of combinations
Let A = {a, b, c}
We want count
the subsets of 2 elements
which we can extract
from the set with 3 elements.
In other words, we want count
the 2-possibilities of combinations
(possible combinations with 2 elements)
which we can extract
from the set with 3 elements.
Enzo Exposyto 33

34. Factorial - 13
nk k-possibilities of combinations
The 6 possible subsets of 2 elements are:
{a, b}, {a, c}, (1st element: a)
{b, a}, {b, c}, (1st element: b)
{c, a}, {c, b} (1st element: c)
The 6 possible subsets of 2 elements
are named 2-possibilities of combinations.
The symbol of k-possibilities of combinationsis nk
and, in this example, nk = 32
The formula of nk, where n ≥ k > 0, with n = 3 and k = 2, is:
= 32 = 3(3-2+1) = 6nk
= n(n − 1)(n − 2)⋯(n − k + 1)
Enzo Exposyto 34

35. Factorial - 14
nk from k-possibilities of combinations to k-combinations
Note that some subsets
have the same elements:
{a, b} and {b, a}
{a, c} and {c, a}
{b, c} and {c, b}
and, then, any subset with its “mate”
represent an identical subset:
the not identical subsets are 3.
Therefore,
the possibilities of combinations of 2 elements are 6
the combinations of 2 elements are 3.
Enzo Exposyto 35

36. Factorial - 15
nk from k-possibilities of combinations to k-combinations
How can we calculate, by a formula,
the number of combinations of 2 elements?
If we note that any subset of 2 elements
has 2 possible permutations,
as the subset {a, b} and its “mate” {b, a},
and that the number of permutations,
for any subset of 2 elements, is given by k! = 2! (page 15),
we must divide the possibilities of combinations (nk = 6)
by the permutations (k! = 2) of a subset of 2 elements.
The result is 3,
that is the number of combinations of 2 elements.
Enzo Exposyto 36

37. Factorial - 16
nk from k-possibilities of combinations to k-combinations
What is the meaning of the number
of combinations of 2 elements?
It's the number of not identical subsets of 2 elements
which we can extract from the set of 3 elements.
In other words, from the set
A = {a, b, c}
we get 3 subsets:
{a, b}, {a, c}, {b, c}
Also, the 3 subsets are 3 combinations of 2 elements.
The number 3 is given by the formula, with n ≥ k > 0:
= = 3(3-2+1) = 6 = 3
k! k(k-1)(k-2)…(1) (2)(1) 2
nk
n(n − 1)(n − 2)⋯(n − k + 1)
Enzo Exposyto 37

38. BINOMIAL
COEFFICIENT
Enzo Exposyto 38

39. BINOMIAL COEFFICIENT - 1
binomial coefficient - 1st formula
It's given by the formula (see page 37)
=
where n ≥ k > 0.
Note that, if k = 0, we can't calculate the numerator:
the last factor (n-k+1) becomes n+1 which is greater than n.
The binomial coefficient represents
the number of distinct combinations with k elements
(or the number of not identical subsets with k elements)
which we can extract
from a set with n elements.
(
n
k)
(
n
k)
nk
k!
=
n(n − 1)(n − 2)⋯(n − k + 1)
k(k − 1)(k − 2)⋯(1)
Enzo Exposyto 39

40. BINOMIAL COEFFICIENT - 2
binomial coefficient - 1st formula - 1st example
=
= 10
(
n
k)
(
n
k)
nk
k!
=
n(n − 1)(n − 2)⋯(n − k + 1)
k(k − 1)(k − 2)⋯(1)
(
5
3)
=
5(4)(5 − 3 + 1)
3!
=
5(4)(3)
6
=
60
6
Enzo Exposyto 40

41. BINOMIAL COEFFICIENT - 3
binomial coefficient - 1st formula - 2nd example
=
= 35
(
n
k)
(
n
k)
nk
k!
=
n(n − 1)(n − 2)⋯(n − k + 1)
k(k − 1)(k − 2)⋯(1)
(
7
3)
=
7(6)(7 − 3 + 1)
3!
=
7(6)(5)
6
=
210
6
Enzo Exposyto 41

42. BINOMIAL COEFFICIENT - 4
binomial coefficient - 2nd formula
Because
then
(
n
k)
[nk
] ⋅ [(n − k)!] = [n(n − 1)⋯(n − k + 1)] ⋅ [(n − k)⋯(1)]
= n!
nk
⋅ (n − k)! = n!
Enzo Exposyto 42

43. BINOMIAL COEFFICIENT - 5
binomial coefficient - 2nd formula
Now, if we multiply numerator and denominator of
by (see previous page), we get:
and, then
with n ≥ k ≥ 0
(
n
k)
nk
k!
(n − k)!
nk
k!
=
nk
⋅ (n − k)!
k!(n − k)!
=
n!
k!(n − k)!
(
n
k)
=
n!
k!(n − k)!
Enzo Exposyto 43

44. BINOMIAL COEFFICIENT - 6
binomial coefficient - 2nd formula - 1st example
= 10
(
n
k)
(
n
k)
=
n!
k!(n − k)!
(
5
3)
=
5!
3!(5 − 3)!
=
5!
3!(2)!
=
120
6 ⋅ (2)
=
120
12
Enzo Exposyto 44

45. BINOMIAL COEFFICIENT - 7
binomial coefficient - 2nd formula - 2nd example
= 35
(
n
k)
(
n
k)
=
n!
k!(n − k)!
(
7
3)
=
7!
3!(7 − 3)!
=
7!
3!(4)!
=
5040
6 ⋅ (24)
=
5040
144
Enzo Exposyto 45

46. BINOMIAL COEFFICIENT - 8
binomial coefficient - 2nd formula - 3rd example
The binomial coefficient occurs in the binomial formula
where n ≥ k. This explains its name.
In this formula, we use the binomial coefficient given by
where n ≥ k ≥ 0
(
n
k)
(x + y)n
=
n
∑
k=0
(
n
k)
xn−k
yk
(
n
k)
=
n!
k!(n − k)!
Enzo Exposyto 46

47. BINOMIAL COEFFICIENT - 9
binomial coefficient - 2nd formula - 3rd example
For example (x ≠ 0 and y ≠ 0):
= + +
= 2! + 2! + 2!
0!2! 1!1! 2!0!
= 2 + 2 + 2 [ = 1 and = 1]
2 1 2
= + 2 +
(
n
k)
(x + y)2
=
2
∑
k=0
(
2
k)
x2−k
yk
(
2
0)
x2−0
y0
(
2
1)
x2−1
y1
(
2
2)
x2−2
y2
x2
y0
x1
y1
x0
y2
x2
y0
x1
y1
x0
y2
y0
x0
x2
xy y2
Enzo Exposyto 47

48. BINOMIAL COEFFICIENT - 10
binomial coefficient - 3rd formula - 1st example
(
n
k)
(
n
k)
=
nk
k!
=
k
∏
i=1
n + 1 − i
i
(
5
3)
=
3
∏
i=1
5 + 1 − i
i
=
5 + 1 − 1
1
⋅
5 + 1 − 2
2
⋅
5 + 1 − 3
3
=
5
1
⋅
4
2
⋅
3
3
= 5 ⋅ 2 ⋅ 1 = 10
Enzo Exposyto 48

49. BINOMIAL COEFFICIENT - 11
binomial coefficient - 3rd formula - 2nd example
(
n
k)
(
n
k)
=
nk
k!
=
k
∏
i=1
n + 1 − i
i
(
7
3)
=
3
∏
i=1
7 + 1 − i
i
=
7 + 1 − 1
1
⋅
7 + 1 − 2
2
⋅
7 + 1 − 3
3
=
7
1
⋅
6
2
⋅
5
3
= 7 ⋅ 3 ⋅
5
3
= 35
Enzo Exposyto 49

50. TETRATION
Enzo Exposyto 50

51. TETRATION
example of tetration - Rucker notation
It means:
Exponentiation tower is builded by 4 same elements (base 2): on the base of tetration
there are 1 exponent and, above, 2 exponents of exponent. It must be evaluated from
top to bottom (or right to left) and it isn't equal to
4
2
4
2 = 2222
= 2[
2(22
)
]
= 2(24
) = 216
= 65,536
4
2
4
2 ≠ [(22
)
2
]
2
= 2(2⋅2⋅2)
= 28
= 256
Enzo Exposyto 51

52. TETRATION
example of tetration
It means:
Exponentiation tower is builded by 3 same elements (base 3): on the base of tetration
there are 1 exponent and, above, 1 exponent of exponent. It must be evaluated from top
to bottom (or right to left) and it isn't equal to
3
3
3
3 = 333
= 3(33
) = 327
3
3
3
3 ≠ (33
)
3
= 3(3⋅3)
= 39
Enzo Exposyto 52

53. TETRATION
tetration
It means:
Exponentiation tower is builded by n same elements (base b):
on the base of tetration there are 1 exponent
and, above, (n-2) exponents of exponent.
It must be evaluated from top to bottom (or right to left).
n
b
n
b = bb⋅⋅⋅b
⏟
n
n
b
Enzo Exposyto 53

54. PENTATION
Enzo Exposyto 54

55. Pentation
is the operation of repeated tetration,
just as
Tetration
is the operation of repeated exponentiation.
Enzo Exposyto 55

56. PENTATION
example of pentation
It means:
a power tower of height 4 2-elements
It is because the exponents in parentheses,
on the left side, mean:
2
2
2
2
2
2 = (
2
2)2 = 4
2 = 2222
⏟
4
= 224
= 216
= 65,536
2
2 = 22
= 4
Enzo Exposyto 56

57. PENTATION
example of pentation
It means:
a power tower of height 7,625,597,484,987 3-elements
This is because the exponents in parentheses,
on the left side, mean:
3
3
3
3
3
3 = (
3
3)3 = 7,625,597,484,987
3 = 333⋅⋅⋅3
⏟
7,625,597,484,987
3
3 = 333
= 3(33
) = 327
Enzo Exposyto 57

58. PENTATION
example of pentation
It means:
a power tower of height b-elements
b
b
b
b
b
b = (
b
b)b = bb⋅⋅⋅b
⏟
b
b
b
b
Enzo Exposyto 58

59. SitoGraphy
Enzo Exposyto 59

60. https://mathmaine.wordpress.com/2010/04/01/sigma-and-pi-notation/
http://www.columbia.edu/itc/sipa/math/summation.html
http://en.m.wikipedia.org/wiki/Tetration
http://mathworld.wolfram.com/PowerTower.html
http://en.m.wikipedia.org/wiki/Pentation
Enzo Exposyto 60