Download Pascal`s Triangle

PASCAL’S TRIANGLE
𝑛
We defined the number ( ) of r-subsets of a set of n elements and
𝑟
showed that
𝑛
( )=
𝑟
𝑛!
𝑟! ∙ (𝑛−𝑟)!
, n, r ∈ 𝕎, r ≤ 𝑛
We write these numbers in the following way.
If we evaluate each entry, we obtain the following triangular array of
numbers.
Value of n
0
1
1
1
1
2
1
2
1
3
1
3
3
1
4
1
4
6
4
1
5
1
5
10
10
5
1
6
1
6 15
20
15 6
1
...
...
This array is known as Pascal’s Triangle. Notice that in any row of
Pascal’s Triangle, entries equidistant (the same distance) from each end
are equal. For example,
That is,
5
5
( )= ( )=1
0
5
5
5
( )= ( )=5
1
4
5
5
( ) = ( ) = 10
3
2
and
6
6
( )= ( )=1
0
6
6
6
( )= ( )=6
1
5
6
6
( ) = ( ) = 15
2
4
The generalization of this property is
𝑛
𝑛
( )= (
)
𝑛−𝑟
𝑟
Proof
𝑛
(
)=
𝑛−𝑟
=
𝑛!
(𝑛−𝑟)! ∙ [𝑛−(𝑛−𝑟)]!
𝑛!
(𝑛−𝑟)! ∙ 𝑟!
𝑛
=( )
𝑟
Example
𝑛
To what other expression in the form ( ) in each of the following
𝑟
equal?
9
17
43
8
(a) ( )
(b) ( )
(c) ( )
(d) ( )
3
6
19
𝑥
Solution
(a) In this case n = 9 and r = 3, so
9
9
9
( )= (
)=( )
3
9−3
6
17
17
17
(b) ( ) = (
)=( )
6
17 − 6
11
43
43
43
(c) ( ) = (
)=( )
19
43 − 19
24
8
8
(d) ( ) = (
)
𝑥
8−𝑥
Example
7
The calculations of ( ) for r = 0, 1, 2, 3, ... , 7 can be simplified by
𝑟
using this property.
7!
7
( )=
=1
0
0! ∙ 7!
7!
7
( )=
=7
1
1! ∙ 6!
7!
6 ∙7
7
( )=
=
= 21
2
2! ∙ 5! 1 ∙ 2
7!
5∙6 ∙7
7
( )=
=
= 35
3
3! ∙ 4! 1 ∙ 2 ∙ 3
7
7
( ) = ( ) = 35
3
4
7
7
( ) = ( ) = 21
5
2
7
7
( )=( )= 7
6
1
7
7
( )=( )= 1
7
0
Notice that each number in any row of Pascal’s Triangle can be obtained
by adding the numbers to the left and right of it in the row above.
This makes it a simple matter to extend Pascal’s Triangle to further
rows.
Value of n
0
1
1
1
1
2
1
2
1
3
1
3
3
1
4
1
4
6
4
1
5
1
5
10
10
5
1
6
1
6 15
20
15 6
1
7
1
7
21
35
35 21
7
1
...
...
Example
(a) Extend Pascal’s Triangle to include the row where n = 8 by using the
pattern illustrated.
8
(b) Check these values by evaluating ( ) for r = 0, 1, 2, ... , 8.
𝑟
Solution
(a)
Value of n
0
1
1
1
1
2
1
2
1
3
1
3
3
1
4
1
4
6
4
1
5
1
5
10
10
5
1
6
1
6 15
20
15 6
1
7
1
7
21
35
35 21
7
1
8
1
8 28
56
70
56
28 8
1
...
...
8!
8
( )=
=1
0
0! ∙ 8!
8!
8
( )=
=8
1
1! ∙ 7!
8!
7 ∙8
8
( )=
=
= 28
2
2! ∙ 6! 1 ∙ 2
8!
6∙7 ∙8
8
( )=
=
= 56
3
3! ∙ 5! 1 ∙ 2 ∙ 3
8!
5∙6∙7 ∙8
8
( )=
=
= 70
4
4! ∙ 4! 1 ∙ 2 ∙ 3 ∙ 4
8
8
( ) = ( ) = 56
5
3
8
8
( ) = ( ) = 28
6
2
8
8
( )=( )= 8
7
1
8
8
( )=( )= 1
8
0
Now we investigate what property of subsets of sets is suggested by this
pattern.
This pattern discovered suggests that
4
3
3
( )+ ( )=( )
1
2
2
4
4
5
( )+ ( )=( )
3
2
3
6
5
5
( )+ ( )=( )
2
1
2
𝑛
Example What expression in the form ( ) would we predict to be equal
𝑟
to each of the following?
7
7
23
23
(a) ( ) + ( )
(c) ( ) + ( )
5
4
18
19
9
64
64
9
(b) ( ) + ( )
(d) ( ) + ( )
2
3
30
29
Solution
7
7
8
(a) ( ) + ( ) = ( )
5
4
5
9
10
9
(b) ( ) + ( ) = ( )
2
3
3
23
23
24
(c) ( ) + ( ) = ( )
18
19
19
64
64
65
(d) ( ) + ( ) = ( )
30
29
30
The generalization of this pattern is known as Pascal’s Theorem.
PASCAL’S THEOREM
𝑛
𝑛
𝑛+1
( )+ (
)= (
)
𝑟
+
1
𝑟
𝑟+1
Proof
𝑛
𝑛
( )+(
)=
𝑟+1
𝑟
=
=
=
=
=
=
=
𝑛!
𝑛!
(𝑟+1)![𝑛−(𝑟+1)]!
𝑟!(𝑛−𝑟)!
𝑛!
𝑛!
+ (𝑟+1)![𝑛−(𝑟+1)]!
𝑟!(𝑛−𝑟)!
𝑛!
𝑛!
+ (𝑟+1)!(𝑛−𝑟−1)!
𝑟!(𝑛−𝑟)!
𝑛!(𝑟+1)
𝑛!(𝑛−𝑟)
+ (𝑟+1)!(𝑛−𝑟)(𝑛−𝑟−1)! (Common Denominator)
(𝑟+1)𝑟!(𝑛−𝑟)!
𝑛!(𝑟+1)+ 𝑛!(𝑛−𝑟)
+
(𝑟+1)!(𝑛−𝑟)!
𝑛![(𝑟+1)+(𝑛−𝑟)]
(𝑟+1)!(𝑛−𝑟)!
𝑛!(𝑛+1)
(Common Factor)
(𝑟+1)!(𝑛−𝑟)!
(𝑛+1)!
(𝑟+1)![(𝑛+1)− (𝑟+1)]!
=(
𝑛+1
).
𝑟+1