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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!
n Cr    
 r  r !(n  r )!
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 n
 
r
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Evaluating binomial coefficient
0011 0010 1010 1101 0001 0100 1011
• Example
 n
n!
n Cr  

 r  r !(n  r )!
 6
6!
6!
6  5  4  3  2 1


 15
 
 2  2!(6  2)! 2!4! 2 1  4  3  2 1
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 8
8!
8!
8!



1
 
0
0!(8

0)!
0!8!
1

8!
 
Your Turn
0011 0010 1010 1101 0001 0100 1011
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 n
n!
n Cr    
 r  r !(n  r )!
Answer
0011 0010 1010 1101 0001 0100 1011
5
5!
5  4  3  2 1


 
2 1 3  2 1
 2  2! 5  2  !
20

 10
2
 n
n!
n Cr    
 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
1
 x  a
1
 x  a
2
 xa
 x 2  2ax  a 2
 x  a   x3  3ax 2  3a 2 x  a3
3
 x  a
4
 x  a
 x 4  4ax3  6a 2 x 2  4a 3 x  a 4
5
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 x5  __ ax4  __ a2 x3  __ a3 x2  __ a 4 x  a5
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
4
=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
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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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