Download Pascal`s Triangle

Chapter 10
Lesson
10-2
Vocabulary
Pascal’s Triangle
Pascal’s Triangle
(r + 1)st term in row n
BIG IDEA
Pascal’s Triangle displays numbers of combinations
in an easy-to-use way.
Suppose you toss a coin five times. As you saw in Lesson 10-1, you
can determine the number of possible arrangements of heads and tails
using combinations. A systematic consideration of the problem leads to
Pascal’s Triangle, an array of numbers that has remarkable properties
and applications in different areas of mathematics.
Mental Math
Expand.
a. (a – b)2
b. (a – 2b)2
c. (3a – 2b)2
Activity
d. (3a2 – 2b)2
Consider five tosses of a coin. Follow the steps below to complete the table
with a count of the number of possible arrangements of heads.
Heads
0
1
2
3
4
5
Total
Arrangements
?
?
?
?
?
?
32
nC r
5C 0
5C 1
?
?
?
?
Step 1 First, suppose there are 0 heads. Then there are 5 tails. There is only
one possible arrangement, which is shown below.
Write this number in the table above. Notice that 1 = 5C0.
Step 2 If there is 1 head, then there are 4 tails. The 1 head can be in any one
of the five places, so there are 5 possible arrangements. Record this
number in the table. Notice that 5 = 5C1. This makes sense, as we are
choosing one position from the five as the location of the 1 head.
Step 3 If there are 2 heads, then there are 3 tails. How many arrangements
are there with 2 heads? Write your answer in the table and explain why
this can be calculated as 5C2.
Step 4 Continue with 3 heads, 4 heads, and 5 heads. Fill in the remaining
entries in the table.
Step 5 To check your work, add the numbers in the second row of the table.
There should be 32 arrangements in all. Each coin can be either
heads or tails, so by the Multiplication Counting Principle, there are
2 · 2 · 2 · 2 · 2 = 25 = 32 possible arrangements.
624
Binomial Distributions
SMP_SEFST_C10_L02_624_630_FINAL.624 624
6/9/09 3:05:20 PM
Lesson 10-2
What Is Pascal’s Triangle?
To count all the possible arrangements of n coins, you can calculate nC 0,
nC 1, nC 2, ..., nC n, a total of n + 1 calculations, and find the sum of those
values. For instance, for n = 4, the 5 values are
4C 0
=1
4C 1
=4
4C 2
=6
4C 3
=4
=1
4C 4
and the total number of arrangements is 1 + 4 + 6 + 4 + 1 = 16.
When the values of nCr are displayed systematically, a beautiful pattern
emerges. At the left below, the values of nCr with values of n and r from
0 to 6 are arranged in an array in the form of a right triangle. At the
right below is the more common isosceles triangle arrangement.
(right triangle format)
n
Pascal’s Triangle
(isosceles triangle format)
nC r
0
0
1
1
1
1
2
1
2
1
3
1
3
3
4
1
4
6
4
1
5
1
5
10
10
5
1
6
1
6
15
20
15
6
1
2
3
4
5
6
row
1
0
1
1
1
2
1
1
3
1
4
1
5
1
6
1
2
3
4
5
6
1
6
10
15
1
3
1
4
10
20
1
5
15
1
6
In the Western world, the array shown in two ways above is called
Pascal’s Triangle, after Blaise Pascal, a French mathematician and
philosopher. Notice that the row numbering starts with row 0 at the top.
1
Blaise Pascal (1623-1662)
Because entries in Pascal’s Triangle can be identified by the row
number n and column number r, the array can be considered as a twodimensional sequence. The following definition provides an explicit
formula for the terms in the nth row of Pascal’s Triangle, where n can
be any whole number. It is an algebraic definition for the triangle.
Definition of Pascal’s Triangle
Let n and r be integers with 0 ≤ r ≤ n. Pascal’s Triangle is the
two-dimensional sequence in which the (r + 1)st term in row n
is nCr or ( nr ).
Example 1
Find the first four terms in row 9 of Pascal’s Triangle.
Solution By the definition of Pascal’s Triangle, the first term of row 9 is 9C 0,
the second term is 9C1, the third term is 9C 2, and the fourth term is 9C 3. From
9!
9!
9!
9!
the formula for C , these are _, _, _, and _ or 1, 9, 36, and 84.
n r
9!0! 8!1! 7!2!
6!3!
Pascal’s Triangle
SMP_SEFST_C10_L02_624_630_FINAL.625 625
625
6/9/09 3:06:46 PM
Chapter 10
QY1
QY1
Properties of Pascal’s Triangle
Looking closely at Pascal’s Triangle, you can find many patterns
and sequences in the rows, columns and diagonals. Here are some
properties that can be proved true for every row of Pascal’s Triangle.
The properties are described both in words and in symbols.
In which row of Pascal’s
Triangle is 10C6? Which
term is 10C6 in that row?
1. The first and last terms in each row are ones. That is, for each
whole number n, nC 0 = nCn = 1.
2. The second and next-to-last terms in the nth row equal n. For each
whole number n, nC 1 = nCn–1 = n.
3. Each row is symmetric. For any whole number n, nCr = nCn–r .
4. The sum of the terms in row n is 2n. For any whole number n,
n
∑
nC r
= 2 n.
r=0
QY2
QY2
Because the entries in Pascal’s Triangle can be interpreted in terms of
choosing or arranging objects, you have already seen some justifications
for these properties based on counting principles. It is also possible to
n!
give formal algebraic proofs using the definition nCr = _
. You are
(n - r)!r!
asked to verify Property 1 in Question 6. Properties 2 and 3 are proved
in Examples 2 and 3, respectively. In the next lesson you are asked to
prove Property 4.
a. Give the first two
and last two terms
of row 95 of Pascal’s
Triangle.
b. Find the sum of all
the entries in row 9
of Pascal’s Triangle.
If you do not understand the statement you are asked to prove, it may
help to use a specific example. For instance, in Property 2, let n = 5.
Then the statement to be proved is that the 2nd and 5th terms in row 5
equal 5, that is, that 5C 1 = 5C 5–1 = 5. From the formula for nCr,
5!
5!
C =_
= 5 and C = C = _
= 5.
5
1
5
4!1!
5–1
5
4
1!4!
Example 2 generalizes the specific example computed above.
Example 2
Prove that the second and next-to-last terms in the nth row of Pascal’s Triangle
equal n.
Solution In any row n, the second entry is nC1. By the formula for nCr,
n!
n!
C = _ = _ = n.
n 1
(n - 1)!1!
(n - 1)!
The next-to-last entry in the nth row is nCn-1. Using the nCr formula again,
n!
n!
n!
C
= __ = __ = _ = n.
n n-1
(n - (n - 1))!(n - 1)!
So, nC1= nCn-1 = n.
626
(n - n + 1)!(n - 1)!
1!(n - 1)!
Binomial Distributions
SMP_SEFST_C10_L02_624_630_FINAL.626 626
6/16/09 10:55:42 AM
Lesson 10-2
The proof of Example 2 can be broadened to prove Property 3, which is
so important that we label it as a theorem.
Theorem (Pascal Triangle Symmetry)
If n and r are any whole numbers with n ≥ r, nCr = nCn - r .
Proof Use the Formula for nCr Theorem.
n!
_
nCr = (n - r)!r!
n!
= _______________
(n - (n - r))!(n - r)!
nCn-r
n!
= ______________
(n - n + r)!(n - r)!
n!
_
=
r!(n - r)!
So for whole numbers n and r, where r ≤ n, nCr = rCn-r .
QY3
QY3
A fi fth property of Pascal’s Triangle gives a recursion relation for
the two-dimensional array. It shows how each row is related to the
preceding row.
5. Each element in Pascal’s Triangle is the sum of the two elements
What does the Pascal
Triangle Symmetry Theorem
say when n = 17 and
r = 11?
nearest it in the preceding row. Specifically, for whole numbers n
and r with 1 ≤ r ≤ n, nCr–1 + nCr = n+1Cr .
For example, the first 4 and 6 in row 4 add to 10, which is the entry just
below these numbers in row 5, as shown below.
row n
0
1
1
1
2
1
3
1
4
1
1
5
6
1
2
3
4
5
6
1
6
10
15
1
3
1
4
10
20
1
5
15
1
6
1
In Question 14, you are asked to give an algebraic proof of this property.
Here is an argument using combinations that shows why the property
works for a specific case. Suppose you wished to know 7C 3, the number
of combinations of 7 objects taken 3 at a time. Call the objects A, B, C,
D, E, F, and G. Each of the 7C 3 combinations either contains G or it
doesn’t. The combinations that contain G include 2 objects from A–F, so
6C 2 combinations contain G. The number of combinations that do not
contain G is the number of ways to pick 3 objects from A–F, so 6C 3 do
not contain G. In this way, 7C 3 = 6C 2 + 6C 3.
Pascal’s Triangle
SMP_SEFST_C10_L02_624_630_FINAL.627 627
627
6/15/09 10:42:10 AM
Chapter 10
GUIDED
Example 3
Use Properties 1, 3, and 5 to construct row 7 in Pascal’s Triangle.
Solution Start by copying row 6.
row 6
1
6
?
?
?
?
?
For row 7, Property 1 states that its first and last entries are 1. To find the
second, third, and fourth entries in row 7, add consecutive pairs of terms in
row 6 as described in Property 5; the results are 1 + 6 = 7, 6 + 15 = ? ,
and 15 + 20 = ? . Use the Pascal Triangle Symmetry Theorem (or continue
adding pairs of terms of row 6) to find the rest of the terms. Copy
the pattern below and write in the result.
?
?
?
?
?
row 6
1
6
row 7
1
Check 1
7
?
?
?
?
?
1
Do the entries equal 7C0, 7C1, 7C2, … ? Check with a calculator.
Check 2 Use Property 4. Does the sum of the elements equal 27 = 128?
Yes, it checks.
A Brief History of Pascal’s Triangle
The array we call “Pascal’s Triangle” was actually discovered many
centuries before Pascal. The Persian mathematician and poet Omar
Khayyam (c. 1048–1122) used the triangle of numbers around the
year 1100. The Chinese mathematician Chu Shih-Chieh wrote about
the array in books published in 1299 and 1303. Perhaps the earliest
description of the triangle comes from the Jain poet and mathematician
Pingala, who lived in India about 200 B.C.E. and wrote in Sanskrit, the
language of sacred Hindu and Jain texts. Pingala discovered methods
for calculating the entries in Pascal’s Triangle, and referred to the array
as the meruprastara, or “Tower of Jewels.”
The first Western discoverer of the triangle was Pascal himself, who
wrote extensively about this triangular array of numbers and its
properties in a 1653 publication, Treatise on the Arithmetic Triangle.
It is for this reason that Westerners refer to the triangle by his name.
Questions
COVERING THE IDEAS
1. a. How many terms are in row 6 of Pascal’s Triangle?
b. The fi fth term is 6Cr . What is the value of r? Of 6Cr?
c. The middle term is 6C s . What is the value of s? Of 6C s?
d. What is the sum of the numbers in this row? Express your
answer as a power of 2.
628
Binomial Distributions
SMP_SEFST_C10_L02_624_630_FINAL.628 628
6/15/09 10:44:17 AM
Lesson 10-2
2. In how many ways can six coins be arranged with two heads and
four tails?
In 3–5, match the English description of the pattern in Pascal’s Triangle to the
description using combination notation.
A For each n, ( 1n ) = (n -n 1 ) = n.
n .
B For each n and each r (r ≤ n), ( nr ) = (n r)
n
n
C For each n, ( 0 ) = (n ) = 1.
3. Each row in the isosceles triangle is symmetric to a vertical line.
4. The second and next-to-last entries in each row are equal to the
row number.
5. The first and last entries in a row are 1.
6. Prove that for all integers n ≥ 1, nC 0 = nCn = 1.
7. Let n = 11, r = 7.
a. Compute nCr-1 + nCr and n+1Cr.
b. Which property stated in this lesson is illustrated by Part a?
8. a. Give the entries in row 8 of Pascal’s Triangle.
b. Check your answer by showing that the entries add to 28.
9. The first nine entries of row 16 of Pascal’s Triangle are 1, 16, 120,
560, 1820, 4368, 8008, 11440, 12870.
a. How many other entries are there in row 16?
b. Without doing any calculations, list the remaining entries.
c. What is the sum of the entries in row 16?
d. List the first four entries in row 17.
10. What is the 10th term in row 20 of Pascal’s Triangle?
APPLYING THE MATHEMATICS
11. What are the last two terms in the row of Pascal’s Triangle whose
terms add to 234?
12. a. Expand the following.
i. (x + y)1
ii. (x + y)2
iii. (x + y)3
b. Relate the coefficients in Part a to Pascal’s Triangle.
c. Verify your conjecture by expanding (x + y)6 on a CAS.
13. The arrows at the right indicate
the diagonals of Pascal’s Triangle.
a. What is the nth term in the
first diagonal?
b. What is the nth term in the
second diagonal?
c. What is the nth term in the
third diagonal?
1st
2nd
3rd
1
4th
1
1
1
1
1
3
4
5
1
2
1
3
6
10
1
4
10
1
5
1
Pascal’s Triangle
SMP_SEFST_C10_L02_624_630_FINAL.629 629
629
6/9/09 3:09:43 PM
Chapter 10
14. Complete Parts a–c to prove Property 5 in the lesson, which states
that any entry in Pascal’s Triangle (presented in an isosceles
triangle array) is the sum of the two entries closest to it in the
row above it. That is, show that for all positive integers r and n
with 1 ≤ r ≤ n, n+1Cr = nCr–1 + nCr .
a. Write an expression for nCr–1 + nCr .
b. Use a CAS to rewrite the expression from Part a with a
single denominator.
c. Explain why the expression in Part b is equivalent to n+1Cr .
15. Suppose that five fair coins are tossed. As in the Activity,
the number of arrangements of 0 heads is 1, the number of
arrangements of 1 head is 5, and so on.
a. Suppose that each arrangement of heads and tails is equally
1
likely. Explain why the probability of each arrangement is _
.
32
5
.
b. Explain why the probability of getting 1 head in 5 tosses is _
32
c. Find the probability of getting 2 heads in 5 tosses.
REVIEW
16. An 18-team softball league split into two divisions of 9 teams each.
In how many ways can this be done? (Lesson 10-1)
17. A 20-person law firm, managed by 4 senior partners, holds a
random drawing in which 4 employees each get a laptop
computer. What is the probability that the four employees
chosen are the 4 senior partners? (Lesson 10-1)
In 18 and 19, solve for x. (Lesson 9-4)
18. 0.3x = 0.07
19. 4(0.2)x = 0.9
20. Skill Sequence Find all real solutions. (Lesson 7-7)
a. x4 + x 3 = 0
b. x4 + x 3 - x - 1 = 0
EXPLORATION
21. Although Pascal is most often associated with the triangle in this
lesson, he had many other interests in science, philosophy, and
religion. Research those interests and describe one of his ideas.
QY ANSWERS
1. 7th term in row 10
2. a. 1, 95, 95, and 1
b. 29 = 512
3.
630
17C11
= 17C6
Binomial Distributions
SMP_SEFST_C10_L02_624_630_FINAL.630 630
6/9/09 3:09:53 PM