The document discusses binomial coefficients and the binomial theorem. It provides examples of expanding binomial expressions using Pascal's triangle and the binomial coefficient formula. The key points are: 1) Binomial coefficients describe the coefficients that arise when expanding binomial expressions using the binomial theorem. 2) Pascal's triangle provides a visual representation of the binomial coefficients. 3) The binomial theorem can be used to expand binomial expressions in terms of binomial coefficients and write individual terms.

1. The Binomial & Multinomial
Coefficients

2.  In formulas arising from the analysis of
algorithms in computer science, the binomial
coefficients occur over and over again, so
that a facility for manipulating them is a
necessity.
 Moreover, different approaches to problems
often give rise to formulas that are different
in appearance yet identities of binomial
coefficients reveal that they are, in fact, the
same expressions.

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The binomial theorem provides a useful method for raising any
binomial to a nonnegative integral power.
Consider the patterns formed by expanding (x + y)n.
(x + y)0 = 1
(x + y)1 = x + y
(x + y)2 = x2 + 2xy + y2
(x + y)3 = x3 + 3x2y + 3xy2 + y3
(x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4
(x + y)5 = x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5
Notice that each expansion has n + 1 terms.
1 term
2 terms
3 terms
4 terms
5 terms
6 terms
Example: (x + y)10 will have 10 + 1, or 11 terms.

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Consider the patterns formed by expanding (x + y)n.
(x + y)0 = 1
(x + y)1 = x + y
(x + y)2 = x2 + 2xy + y2
(x + y)3 = x3 + 3x2y + 3xy2 + y3
(x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4
(x + y)5 = x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5
1. The exponents on x decrease from n to 0.
The exponents on y increase from 0 to n.
2. Each term is of degree n.
Example: The 5th term of (x + y)10 is a term with x6y4.”

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The coefficients of the binomial expansion are called binomial
coefficients. The coefficients have symmetry.
The coefficient of xn–ryr in the expansion of (x + y)n is written
or nCr .
n
r
 
 
 
(x + y)5 = x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5
The first and last coefficients are 1.
The coefficients of the second and second to last terms
are equal to n.
1 1
Example: What are the last 2 terms of (x + y)10 ? Since n = 10,
the last two terms are 10xy9 + 1y10.
So, the last two terms of (x + y)10 can be expressed
as 10C9 xy9 + 10C10 y10 or as xy9 + y10.






9
10






10
10

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The triangular arrangement of numbers below is called Pascal’s
Triangle.
Each number in the interior of the triangle is the sum of the two
numbers immediately above it.
The numbers in the nth row of Pascal’s Triangle are the binomial
coefficients for (x + y)n .
1 1 1st row
1 2 1 2nd row
1 3 3 1 3rd row
1 4 6 4 1 4th row
1 5 10 10 5 1 5th row
0th row1
6 + 4 = 10
1 + 2 = 3

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Example: Use the fifth row of Pascal’s Triangle to generate the
sixth row and find the binomial coefficients , , 6C4 and 6C2 .
6
1
 
 
 
6
5
 
 
 
5th row 1 5 10 10 5 1
6th row
6
0
 
 
 
6
1
 
 
 
6
2
 
 
 
6
3
 
 
 
6
4
 
 
 
6
5
 
 
 
6
6
 
 
 
6C0 6C1 6C2 6C3 6C4 6C5 6C6
= 6 = and 6C4 = 15 = 6C2.
6
1
 
 
 
6
5
 
 
 
There is symmetry between binomial coefficients.
nCr = nCn–r
1 6 15 20 15 6 1

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Example: Use Pascal’s Triangle to expand (2a + b)4.
(2a + b)4 = 1(2a)4 + 4(2a)3b + 6(2a)2b2 + 4(2a)b3 + 1b4
= 1(16a4) + 4(8a3)b + 6(4a2b2) + 4(2a)b3 + b4
= 16a4 + 32a3b + 24a2b2 + 8ab3 + b4
1 1 1st row
1 2 1 2nd row
1 3 3 1 3rd row
1 4 6 4 1 4th row
0th row1

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The symbol n! (n factorial) denotes the product of the first n
positive integers. 0! is defined to be 1.
n! = n(n – 1)(n – 2)  3 • 2 • 1
1! = 1
4! = 4 • 3 • 2 • 1 = 24
6! = 6 • 5 • 4 • 3 • 2 • 1 = 720
Formula for Binomial Coefficients For all nonnegative
integers n and r, !
( )! !
n r
n
C
n r r


Example:
)123()1234(
)123()4567(
••••••
••••••

!3!4
7
!3!4
!7
!3)!37(
!7
•••
37 

C
35
1234
4567
•••
•••


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Example: Use the formula to calculate the binomial coefficients
10C5, 10C0, and .50
48
 
 
 
12
1
 
 
 
!5)!510(
!10
•
510

C
!0)!010(
!10
•
010

C
!48)!4850(
!50
48
50
•






!1)!112(
!12
1
12
•






!5!5
!10
•

!5!5
!5)678910(
•
•••••
 252
12345
678910
••••
••••

!0!10
!10
•
 1
1
1
!0
!1

!48!2
!48)4950(
!48!2
!50
•
••
•
 1225
12
4950
•
•

!1!11
!1112
!1!1
!12
•
•
•
 12
1
12


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Binomial Theorem
1 1
( )n n n n r r n n
n rx y x nx y C x y nxy y  
       L L
!
with
( )! !
n r
n
C
n r r



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Binomial Theorem
Example: Use the Binomial Theorem to expand (x4 + 2)3.
03C 13C 23C 33C 34
)2(x 34
)(x )2()( 24
x 24
)2)((x 3
)2(
1 34
)(x 3 )2()( 24
x 3 24
)2)((x 1
3
)2(
8126 4812
 xxx

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Although the Binomial Theorem is stated for a binomial which
is a sum of terms, it can also be used to expand a difference of
terms.
Simply rewrite
(x + y)n as (x + (– y))n
and apply the theorem to this sum.
Example: Use the Binomial Theorem to expand (3x – 4)4.
4
)43( x 4
))4(3(  x
432234
)4(1)4)(3(4)4()3(6)4()3(4)3(1  xxxx
256)64)(3(4)16)(9(6)4)(27(481 234
 xxxx
25676886443281 234
 xxxx

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Example: Use the Binomial Theorem to write the first three
terms in the expansion of (2a + b)12 .
...)2(
2
12
)2(
1
12
)2(
0
12
)2( 210111212


















 babaaba
...)2(
!2)!212(
!12
)2(
!1)!112(
!12
)2(
!0)!012(
!12 2101011111212
•••






 babaa
...)2)(1112()2(12)2( 2101011111212 •  babaa
...135168245764096 2101112
 babaa

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Example: Find the eighth term in the expansion of (x + y)13 .
Think of the first term of the expansion as x13y0 . The power of
y is 1 less than the number of the term in the expansion.
The eighth term is 13C7 x6y7.
Therefore, the eighth term of (x + y)13 is 1716 x6y7.
!7!6
!7)8910111213(
!7!6
!13
•
••••••
•
713 C
1716
123456
8910111213
•••••
•••••


17. Find the 6th term in the expansion of (3a + 2b)12
Using the Binomial Theorem, let x = 3a and y = 2b
and note that in the 6th term, the exponent of y is
m = 5 and the exponent of x is n – m = 12 – 5 = 7.
Consequently, the 6th term of the expansion is:
57
512 yxC    57
23
!5!7
!789101112
ba


= 55,427,328 a7b5

18. E.g. 7—Finding a Particular Term in a Bin. Expansion
Find the coefficient of x8
in the expansion of
• Both x2 and 1/x are powers of x.
• So, the power of x in each term of the expansion
is determined by both terms of the binomial.
10
2 1
x
x
 
 
 

19. To find the required coefficient, we first
find the general term in the expansion.
• By the formula, we have:
a = x2, b = 1/x, n = 10
• So, the general term is:
E.g. 7—Finding a Particular Term in a Bin. Expansion
10
2 2 1 10
3 10
10 101
( ) ( )
10 10
10
10
r
r r r
r
x x x
r x r
x
r

 

    
         
 
  
 

20. Thus, the term that contains x8
is the term in which
3r – 10 = 8
r = 6
• So, the required coefficient is:
E.g. 7—Finding a Particular Term in a Bin. Expansion
10 10
210
10 6 4
   
    
   