Binomial Coefficient
0011 0010 1010 1101 0001 0100 1011
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Definition of Binomial coefficient
0011 0010 1010 1101 0001 0100 1011
For nonnegative integers n and r with n > r the expansion
(read “n above r”) is called a binomial coefficient and is
defined by
n
n!
Cr    
n
 r  r !( n  r ) !
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n
 
r
2
Evaluating binomial coefficient
0011 0010 1010 1101 0001 0100 1011
• Example
n
Cr
n
n!
   
r !( n  r ) !
r
6
6!
6!
6  5  4  3  2 1



 15
 
2 !(6  2 ) !
2 !4 !
2 1  4  3  2 1
2
1
4
 8 
8!
8!
8!



 1


0
0
!(8

0
)
!
0
!8
!
1

8
!


2
Your Turn
0011 0010 1010 1101 0001 0100 1011
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n
n!
Cr    
n
 r  r !( n  r ) !
Answer
0011 0010 1010 1101 0001 0100 1011
5
5!

 
 2  2 ! 5  2  !

5  4  3  2 1
2 1  3  2 1

20
 10
2
n
n!
Cr    
n
 r  r !( n  r ) !
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Expanding binomial
0011 0010 1010 1101 0001 0100 1011
• The theorem that specifies the expansion of any
power (a+b)n of a binomial (a+b) as a certain sum
of products
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We can easily see the pattern on the x's and the a's.
But what about the coefficients? Make a guess and
then as we go we'll see how you did.
0011 0010 1010 1101 0001 0100 1011
x  a
0
x  a
x  a
1
1
2
 xa
 x  2 ax  a
2
 x  a   x  3 ax  3 a x  a
3
x  a
4
x  a
3
2
1
2
2
3
 x  4 ax  6 a x  4 a x  a
4
5
3
2
2
3
4
4
 x 5  __ ax 4  __ a 2 x 3  __ a 3 x 2  __ a 4 x  a 5
2
Pascal’s Triangle
0011 0010 1010 1101 0001 0100 1011
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Pascal’s Triangle
0011 0010 1010 1101 0001 0100 1011
• Each row of the triangle begins with a 1 and ends with a
1.
• Each number in the triangle that is not a 1 is the sum of
the two numbers directly above it (one to the right and
one to the left.)
• Numbering the rows of the triangle 0, 1, 2, … starting at
the top, the numbers in row n are the coefficients of x n, x
n-1y , x n-2y2 , x n-3y3, … y n in the expansion of (x + y)n.
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Binomial Theorem
0011 0010 1010 1101 0001 0100 1011
• The a’s start out to the nth power and decrease by 1 in
power each term. The b's start out to the 0 power and
increase by 1 in power each term.
• The binomial coefficients are found by computing the
combination symbol. Also the sum of the powers on a and
b is n.
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a+b)n = nCo an bo +nC1 an-1 b1 +nC2 an-2 b2+…..+nCn a0 bn.
(
Example
0011 0010 1010 1101 0001 0100 1011
Write the binomial expansion of (x+y) 7
.
Solution :Use the binomial theorem
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A=x; b=y; n=7
(x+7)7=x7+7c1x6y1+7c2x5y2+7c3x4y3+7c4x3y4+7c5x2y5+
6+ c y7
c
xy
7 6
7 7
Answer
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=x7+7x6y1+21x5y2+35x4y3+35x3y4+21x2y5+7xy6+y7
Question 2
(2x-y) 4
0011 0010 1010 1101 0001 0100 1011
Solution :Use the binomial theorem
a=2x; b=-y; n=y
= (2x) 4=4c1 (2x) 3y+4c2 (2x) 2y2-4c3 (2x)
y3+4c4y4
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Answer
=16x4-32x3y+24x2y2-8xy3+y4
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Question 3
(11)5= (10+1)5
0011 0010 1010 1101 0001 0100 1011
Solution : Use the binomial theorem, to find the value of
A=10; b=1; n=5
=105+5c1104 (1) +5c4103 (1)2+5c3 (10)2(1)3+5c4 (10)5-4(1)4+5c5
(1)
=100000+5x100000+10x1000+5x10+1x1
Answer
=161051.
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GENERAL TERM IN A BINOMIAL
EXPANSION
0011 0010 1010 1101 0001 0100 1011
•
•
•
•
•
•
•
•
For n positive numbers we have
(a+b)n = nCo an bo +nC1 an-1 b1 +nC2 an-2 b2+…..+nCn a0 bn.
According to this formula we have
The first term=T1= nCo an b0
The second term =T2= nC1 an-1 b1
The third term=T3= nC2 an-2 b2
So, any individual terms, let’s say the ith term, in a binomial
Expansion can be represented like this:
Ti=n C(i-1) an-(i-1) b(i-1)
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EXAMPLE
0011 0010 1010 1101 0001 0100 1011
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MIDDLE TERM
0011 0010 1010 1101 0001 0100 1011
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EXAMPLE
0011 0010 1010 1101 0001 0100 1011
• Find the middle term in the expansion of
(4x-y) 8
th term =5th term
Ti=
T5=8C4
(4x)8-4(-y)4
T5= 70(256x4) (y4)
T5=17920x4y4
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Example
0011 0010 1010 1101 0001 0100 1011
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Group Members
0011 0010 1010 1101 0001 0100 1011
• Ayesha Khalid
• Hira Shamim Syed
• Urooj Arshad Syed
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0011 0010 1010 1101 0001 0100 1011
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