Download Uses for Pascal`s Triangle

Uses for Pascal’s Triangle
An entry in Pascal's Triangle is usually given a row number and a place in that row,
beginning with row zero and place, or “element”, zero. For instance, at the top of Pascal's
Triangle is the number 1, which makes up row zero, element zero. The next row, row 1,
contains two 1's, both formed by adding the two numbers above them to the left and the
right, in this case 1 and 0 (all numbers outside the Triangle are 0's). The 2nd row is
created as follows: 0+1=1; 1+1=2; 1+0=1. The third is: 0+1=1; 1+2=3; 2+1=3; 1+0=1.
Likewise, the number 20 appears in row 6, element 3 in Pascal’s Triangle. In this way,
the rows of the triangle go on infinitely.
Patterns & Math in Pascal’s Triangle!
 The Sums of the Rows
Put a token on each of the elements in any one row. Take the sum of the numbers on that
row and write the total as a power of 2. What do you find? (They will be equal to 2 to the
nth power or 2n, when n is the number of the row). For example if you cover all elements
of the 6th row, then add them up, you would get: 1 + 6 + 15 + 20 + 15 + 6 + 1 = 64 = 26
2o = 1
2 = 1+1=2
2
2 = 1+ 2 + 1 = 4
23 = 1 + 3 + 3 + 1 = 8
24 = 1 + 4 + 6 + 4 + 1 = 16
1
Written by Jeanne Lazzarini (RAFT)
Copyright 2011, RAFT
 Prime Numbers
If the 1st element in a row is a prime number then cover that number with a token
(remember, the zeroth element of every row is 1). What do you notice about all the other
numbers in that same row (excluding the 1's)? You should find that all the numbers are
divisible by that prime number: for example, in row 7; 7, 21, and 35 are all divisible by 7.
 Find Any Number in the Triangle!
Any number in the triangle can also be found by nCr (n Choose r) where n is the number
of the row and r is the element in that row. For example, in row 3, 1 is the zeroth element,
3 is the 1st element, the next 3 is the 2nd element, and the last 1 is the 3rd element. The
formula for nCr is:
n!
-------r!(n-r)!
where “!” means “factorial”, or the number multiplied by all the positive integers that are
smaller than the number. For example: 5! = 5 × 4 × 3 × 2 × 1 = 120. For example, ask a
friend to find your number on the board by giving them only the number of the row (in
this case 7) and the element number where 35 appears in that row (which is 3). Using the
formula above, they should find
(7!) ÷ [(3!) (7 – 3)]! =
(7 x 6 x 5 x 4 x 3 x 2 x 1)
number!!
÷ [(3 x 2 x 1) (4 x 3 x 2 x 1)] = 7 x 5 = 35; so 35 was the
 The Probability of Heads or Tails
Pascal originally discovered the properties of this triangle by thinking of problems posed
by gamblers. Investigate what happens when a coin is flipped. If you flip a coin, it will
come up either heads or tails.
In the following analysis the focus is on the probability of flipping heads. An equivalent
could be said for flipping tails. Cover the first 1 at the top of Pascal’s Triangle (row 0)
with a token to represent the flip of zero coins. Next, cover the numbers on the 1st row to
represent the probability of getting a head on the toss of 1 coin. There are only 2
possibilities when tossing one coin and the probability of the coin coming up heads is 1
chance out of 2, or ½. The probability of any event is expressed as a fraction between 0
and 1 inclusive.
Compare the number equivalent outcomes to the numbers in the 1st row of the triangle:
One Coin tossed
1st possible outcome: H
2nd possible outcome: T
Uses for Pascal’s Triangle, page 2
Number of heads
1
0
Number of equivalent outcomes
1
1
Copyright 2011, RAFT
Suppose you toss 2 coins. Now cover the numbers in the 2nd row to represent the number
equivalent outcomes when tossing 2 coins. The following are the possible outcomes:
Look at the number of heads in these possible outcomes:
2nd coin
1st coin
1st possible outcome:
H
H
nd
H
T
2 possible outcome:
3rd possible outcome:
T
H
4th possible outcome:
T
T
Compare the number equivalent outcomes to the numbers in the 2nd row of the triangle:
Number of heads
2
1
0
Number of equivalent outcomes
1
2
1
The probability of getting 2 heads from tossing 2 coins is: 1 chance out of 4 outcomes =
¼. The probability of getting one head and one tail from tossing 2 coins is 2 chances out
of 4 = ½. The probability of getting zero heads (= 2 tails) is one chance out of 4 = ¼.
Taking this further, toss 3 coins, and cover up the 3rd row of numbers. The following
tree diagram represents the possible outcomes:
The probability of getting exactly 3 heads after tossing 3 coins is 1 chance out of 8, or
1/8. The probability of getting exactly 2 heads is 3 chances out of 8, or 3/8.
Uses for Pascal’s Triangle, page 3
Copyright 2011, RAFT
Look at the number of heads in the possible outcomes:
1st coin
2nd coin
1st possible outcome:
H
H
nd
H
H
2 possible outcome:
3rd possible outcome:
H
T
th
4 possible outcome:
H
T
5th possible outcome:
T
H
6th possible outcome:
T
H
th
T
T
7 possible outcome:
8th possible outcome:
T
T
3rd coin
H
T
H
T
H
T
H
T
Compare now the number of heads to the number of equivalent outcomes:
Number of heads
3
2
1
0
Number of equivalent outcomes
1
3
3
1
By now you might recognize that the number of equivalent outcomes are the same as the
numbers in the third row of Pascal’s Triangle!!! If we continue by tossing n coins we
would end up with the number of equivalent outcomes of heads being the same as the
numbers in the nth row of Pascal’s Triangle! For example, if 4 coins are tossed, then the
numbers of equivalent outcomes of heads are the numbers 1 4 6 4 1 which are also
the numbers in the 4th row of Pascal’s Triangle! Predict your odds of getting heads with
coin tossing patterns in Pascal’s Triangle!!!
 The Fibonnacci Sequence!
Fibonnacci's Sequence of numbers starts with 1, 1, and then the next number is added to
the previous number to become the next number in the sequence and so on. This creates:
1,1,2,3,5,8,13,21,34, 55,89,144,233, etc . . . . This sequence can also be located in
Pascal's Triangle! Put a token on the first 1 (in row 0) on the triangle. This represents
marking off that number as the first number in the Fibonnaci sequence. Next, use tokens
to cover up numbers (see illustration below) that appear along each straight line (perhaps
use a different colored playing piece on each straight line).
The numbers in the consecutive rows shown in the diagram
below are the first numbers of the Fibonnacci Sequence. The
Fibonnacci Sequence can be found in many places, including
the Golden Ratio, the lengths of the segments of a pentagram,
the forms of plants and shells, harmonies played on musical
instruments, and more!
Uses for Pascal’s Triangle, page 4
Copyright 2011, RAFT
 Sierpinski's Gasket
Put a token on all the odd numbers in Pascal's Triangle, and leave the rest alone. What
pattern do you see? You will find the recursive Sierpinski Gasketfractal is revealed (see
figure below), yet another fascinating pattern in Pascal's Triangle! Other interesting
patterns are formed if the elements not divisible by other numbers are filled, especially
those indivisible by prime numbers.
For example, cover only the numbers in Pascal’s triangle that are multiples of 3, and then
do this again after clearing the board for multiples of 5, and of 7. What do you see?
See the related RAFT Idea Sheets:
Sierpinski Gasket - http://www.raft.net/ideas/Sierpinski Gasket.pdf
Tetrix - http://www.raft.net/ideas/Tetrix.pdf
 Square Rings
Place a token on each of six numbers surrounding a number on Pascal’s Triangle. What
do you notice? For example (assume shaded squares are covered by tokens):
1 x 1 x 5 x 10 x 6 x 3 = 900 = 302
3 x 4 x 10 x 10 x 4 x 3 = 14,400 = 1202
The product of the surrounding number is a perfect square!
Also the products of alternate triples are equal!
e.g., 1 x 5 x 6 = 1 x 10 x 3 = 30 and 3 x 10 x 4 = 4 x 10 x 3 = 120.
Can you find a relationship between the number that is surrounded and the value of the
squared number?
Uses for Pascal’s Triangle, page 5
Copyright 2011, RAFT
 Figures Formed by Points on a Circle
Study the numbers in the chart below. What do you notice???? The numbers are actually
the top of Pascal's Triangle, except the first 1 (element zero) in each row is missing. The
figures are formed by placing a number of points on the circumference of a circle and
then drawing all the possible lines between them(it does not matter if the points are
equally spaced along the circumference). For instance, for a figure with n points, look at
the nth row of Pascal’s Triangle in order to find the number of points, line segments, and
polygons of the figure formed by having its vertices on the circle! So, if a pentagon is
formed by 5 points (vertices) on the circle, then the 5th row of Pascal’s Triangle shows
the number of points (which is 5), the number of line segments (which are 10), the
number of triangles formed within the figure by line segments (which is another 10), the
number of quadrilaterals formed within the figure by line segments (which are 5), and the
number of pentagons formed (which is one). There are no 6-sided hexagons or 7-sided
heptagons, or any n-sided figures for n < 5.
Uses for Pascal’s Triangle, page 6
Copyright 2011, RAFT
 Binomial Expansions!
Consider the following expanded powers of (a + b)n, where a + b is any binomial and n is
a whole number. Look for patterns!
Unexpanded Form
Expanded Form
(a + b)0 =
1
(a + b)1 =
a + b
(a + b)2 =
a2 + 2ab + b2
(a + b)3 =
a3 + 3a2b + 3ab2 + b3
(a + b)4 =
a4 + 4a3b + 6a2b2 + 4ab3 + b4
(a + b)5 =
a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5
•
• (note: this pattern could go on infinitely!)
•
Each expanded form is a polynomial. Do you see any relationships to Pascal’s Triangle?
1. Notice Pascal’s Triangle is embedded within these polynomial expansions! It turns
out knowing Pascal’s Triangle can help you write the expanded form of a polynomial
without much effort! For example, how can you expand the polynomial (x + 2)3 ?
Well, you could use the technique of multiplying the factors together, which can take
time. But according to Pascal's triangle, in the expanded polynomial for (a + b)3 the
coefficients of each term are 1(from 1a3), 3 (from 3a2b), 3 (from 3ab2), and 1 (from
b3). This means that for (x + 2)3, x is “a”, and 2 is “b”, so by plugging in these
values you get:
(x + 2)3 = x3 + (3x2)(2) + (3x)(22) + 23 =
x3 + 6x2 + 12x + 8, the expanded form!
2. When n is even, (−b)n will be positive. But when n is odd, (−b)n will be negative.
This means that each odd power of b will have a negative sign. Look at the following
example for using Pascal’s triangle to expand (x – 1)5
Again, looking at Pascal's triangle, the binomial coefficients for (a + b)5 are 1 5 10
10 5 1. In this case, x is “a”, and −1 is "b". The signs will alternate, and each odd
power of −1 will have a negative sign :
(x – 1)5 = x5 (-1)0 + 5x4 (-1)1 + 10x3 (-1)2 + 10x2 (-1)3 + 5x (-1)4 + x0 (-1)5
= x5 − 5x4 + 10x3 − 10x2 + 5x – 1, the expanded form!
Uses for Pascal’s Triangle, page 7
Copyright 2011, RAFT
What other ways does Pascal’s Triangle relate to expanded polynomials?
3. For each term in an expanded polynomial, the sum of the exponents is n, the power to
which the binomial is raised. For example, in (a + b) 4, add the exponents in each
term: for a4, which is a4b0, add the exponents 4 + 0 to get 4; for each 4a3b add the
exponents 3 + 1 to get 4; for each 6a2b2 add the exponents 2 + 2 to get 4; and for b4,
which is a0b4, add the exponents 0 + 4 to get 4.
4. Also, notice for every polynomial expansion, there is one more term than the power
of the exponent, n. For example, (a + b) 2 has 3 terms in its expanded form: a2 + 2ab +
b2, which are 1, 2, and 1, and the power of the exponent is 2. For (a + b) 5, there are 6
terms (1, 5, 10, 10, 5, and 1); again one more term than the power of the exponent 5.
5. In each expanded polynomial, the exponents of the first term “a” begin with n, the
power of the binomial, and the power of each successive “a” in each term decreases
by one to the power 0 in the last term. The last term has no factor of a (since a0 = 1).
Likewise, first term has no factor of b (since b0 = 1), and the powers of b start with 0
and increase term by term to the power of n.
6. The coefficients for the terms start at 1 and increase through certain values about
"half"-way and then decrease through these same values back to 1.
Uses for Pascal’s Triangle, page 8
Copyright 2011, RAFT