Transcript MATH 106 Combinatorics

Section 9
Binomial Expansion
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Recall the exercises we did last class:
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What is “Binomial Expansion”?
#1
Take an expression of the form (x + y)n and multiply it out for n = 2 and
for n = 3.
(x + y)2 = (x + y)(x + y) = x2 + xy + yx + y2 = x2 + 2xy + y2
Before simplification, how many terms did we get? 4 = 22
(x + y)3 = (x + y)(x + y)(x + y) =
x3 + xxy + xyx + xyy + yxx + yxy + yyx + y3 = x3 + 3x2y + 3xy2 + y3
Before simplification, how many terms did we get? 8 = 23
RECIPE FOR CHOOSING ONE TERM IN THE EXPANSION OF
(x + y)n BEFORE SIMPLIFICATION:
Choose x or y from first factor and then
choose x or y from second factor and then …
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#2
Without doing any multiplication, find the following expansion:
(x + y)7 = x7 + 7 x6y + 21 x5y2 + 35 x4y3 + 35 x3y4 + 21 x2y5 + 7 xy6 + y7
If we did the multiplication, how many terms would we get before
simplification? 128 = 27
RECIPE FOR CHOOSING ONE TERM IN THE EXPANSION OF
(x + y)n BEFORE SIMPLIFICATION:
Choose x or y from first factor and then
choose x or y from second factor and then …
Observe that the coefficients in the simplified expansion of (x + y)n
match the row of Pascal’s Triangle corresponding to n.
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The Binomial Theorem
(bottom of page 67 in the textbook)
(x + y)n = C(n, 0)xny0 + C(n, 1)xn-1y1 + C(n, 2)xn-2y2 + C(n, 3)xn-3y3 +
… + C(n, n-2)x2yn-2 + C(n, n-1)x1yn-1 + C(n, n)x0yn
Because of symmetry, we could choose to write the formula in the
Binomial Theorem as follows:
(x + y)n = C(n, n)xny0 + C(n, n-1)xn-1y1 + C(n, n-2)xn-2y2 + C(n, n-3)xn-3y3
+ … + C(n, 2)x2yn-2 + C(n, 1)x1yn-1 + C(n, 0)x0yn
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HISTORICAL NOTE:
n
Sometimes the notation C(n, k) is written as  
k 
Both forms are sometimes called the “binomial coefficient.”
Let’s do some!
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#3
Find the expansion of each of the following:
(x + y)4
x4 + 4 x3y + 6 x2y2 + 4 xy3 + y4
(x + 3)4
x4 + 4 x3(3) + 6 x2(3)2 + 4 x(3)3 + (3)4 =
x4 + 12x3 + 54x2 + 108x + 81
(x2 + 3)4
(x2)4 + 4 (x2)3(3) + 6 (x2)2(3)2 + 4 (x2)(3)3 + (3)4 =
x8 + 12x6 + 54x4 + 108x2 + 81
(x – y)4
x4 + 4 x3(–y) + 6 x2(–y)2 + 4 x(–y)3 + (–y)4 =
x4 – 4x3y + 6x2y2 – 4xy3 + y4
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#4
Find the expansion of each of the following:
(x + 3y)4
x4 + 4 x3(3y) + 6 x2(3y)2 + 4 x(3y)3 + (3y)4 =
x4 + 12x3y + 54x2y2 + 108xy3 + 81y4
(x2
–
2)6
(x2)6 + 6 (x2)5(2) + 15 (x2)4(2)2 + 20 (x2)3(2)3 +
15 (x2)2(2)4 + 6 (x2)(2)5 + (2)6 =
x12  12x10 + 60x8  160x6 + 240x4  192x2 + 64
(3x4 + 2y3)5 (3x4)5 + 5 (3x4)4(2y3) + 10 (3x4)3(2y3)2 + 10 (3x4)2(2y3)3 +
5 (3x4)(2y3)4 + (2y3)5 =
243x20 + 810x16y3 + 1080x12y6 + 720x8y9 + 240x8y9 + 32y15
(3x4 – 2y3)5
243x20  810x16y3 + 1080x12y6  720x8y9 + 240x8y9  32y15
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#5
Determine the coefficient of x6y4 in the expansion of (x + y)10
C(10, 4) x6y4 = 210 x6y4
Determine the coefficient of x12y4 in the expansion of (x2 + 2y)10
C(10, 4) (x2)6(2y)4 = 210 (x2)6(2y)4 =
210 x1224y4 = 3360 x12y4
In the expansion of (x4 – 3y3)9 , determine the coefficient of
x20y12
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C( 9, 4 ) ( x4 ) (–3y3 )4 = 126 x20(–3)4y12 = 10206 x20y12
x24y9
C( 9, 3 ) ( x4 ) ( –3y3 ) 3 = 84 x24(–3)3y9 = –2268 x24y9
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Homework Hints:
In Section 9 Homework Problems #3, 4, 5, and 6,
do not actually do the algebraic expansions. Instead, use the
Binomial Theorem and Pascal’s Triangle. Also, don’t forget
that ( y)n is equal to yn if n is even, and ( y)n is equal to  yn
if n is odd.
In Section 9 Homework Problems #7, 8, and 9,
be sure to use Problem #4 on the Section #9 Class Handout
as a guide.
Quiz #2 NEXT CLASS!
Be sure to do the review problems for this, quiz posted on the
internet. The link can be found in the course schedule.
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