The document discusses calculating binomial coefficients without using Pascal's triangle. It begins by stating the formula for calculating binomial coefficients as (n over k)(k factorial) over (n factorial) and proves it using mathematical induction. It then explains how the binomial theorem can be derived from this formula for calculating coefficients. It concludes by providing an example calculation of a binomial coefficient.

1. Calculating Coefficients
without Pascal’s Triangle

2. Calculating Coefficients
without Pascal’s Triangle
 !!
!
knk
n
Ck
n



3. Calculating Coefficients
without Pascal’s Triangle
Proof
 !!
!
knk
n
Ck
n



4. Calculating Coefficients
without Pascal’s Triangle
Proof
 !!
!
knk
n
Ck
n


Step 1: Prove true for k = 0

5. Calculating Coefficients
without Pascal’s Triangle
1
0

 CLHS n
Proof
 !!
!
knk
n
Ck
n


Step 1: Prove true for k = 0

6. Calculating Coefficients
without Pascal’s Triangle
1
0

 CLHS n
Proof
1
!
!
!!0
!



n
n
n
n
RHS
 !!
!
knk
n
Ck
n


Step 1: Prove true for k = 0

7. Calculating Coefficients
without Pascal’s Triangle
1
0

 CLHS n
RHSLHS 
Proof
1
!
!
!!0
!



n
n
n
n
RHS
 !!
!
knk
n
Ck
n


Step 1: Prove true for k = 0

8. Calculating Coefficients
without Pascal’s Triangle
1
0

 CLHS n
RHSLHS 
Proof
1
!
!
!!0
!



n
n
n
n
RHS
 !!
!
knk
n
Ck
n


Step 1: Prove true for k = 0
Hence the result is true for k = 0

9. 0integeraniswherefortrueisresulttheAssume:2Step  rrk

10. 0integeraniswherefortrueisresulttheAssume:2Step  rrk
 !!
!
..
rnr
n
Cei r
n



11. 0integeraniswherefortrueisresulttheAssume:2Step  rrk
 !!
!
..
rnr
n
Cei r
n


1fortrueisresulttheProve:3Step  rk

12. 0integeraniswherefortrueisresulttheAssume:2Step  rrk
 !!
!
..
rnr
n
Cei r
n


1fortrueisresulttheProve:3Step  rk
   !1!1
!
.. 1


rnr
n
Cei r
n

13. 0integeraniswherefortrueisresulttheAssume:2Step  rrk
 !!
!
..
rnr
n
Cei r
n


1fortrueisresulttheProve:3Step  rk
   !1!1
!
.. 1


rnr
n
Cei r
n
Proof

14. 0integeraniswherefortrueisresulttheAssume:2Step  rrk
 !!
!
..
rnr
n
Cei r
n


1fortrueisresulttheProve:3Step  rk
   !1!1
!
.. 1


rnr
n
Cei r
n
Proof
11
1


 r
n
r
n
r
n
CCC

15. 0integeraniswherefortrueisresulttheAssume:2Step  rrk
 !!
!
..
rnr
n
Cei r
n


1fortrueisresulttheProve:3Step  rk
   !1!1
!
.. 1


rnr
n
Cei r
n
Proof
11
1


 r
n
r
n
r
n
CCC
r
n
r
n
r
n
CCC  

 1
1
1

16. 0integeraniswherefortrueisresulttheAssume:2Step  rrk
 !!
!
..
rnr
n
Cei r
n


1fortrueisresulttheProve:3Step  rk
   !1!1
!
.. 1


rnr
n
Cei r
n
Proof
11
1


 r
n
r
n
r
n
CCC
r
n
r
n
r
n
CCC  

 1
1
1
 
     !!
!
!!1
!1
rnr
n
rnr
n






17. 0integeraniswherefortrueisresulttheAssume:2Step  rrk
 !!
!
..
rnr
n
Cei r
n


1fortrueisresulttheProve:3Step  rk
   !1!1
!
.. 1


rnr
n
Cei r
n
Proof
11
1


 r
n
r
n
r
n
CCC
r
n
r
n
r
n
CCC  

 1
1
1
 
     !!
!
!!1
!1
rnr
n
rnr
n





   
   !!1
1!!1
rnr
rnn




18. 0integeraniswherefortrueisresulttheAssume:2Step  rrk
 !!
!
..
rnr
n
Cei r
n


1fortrueisresulttheProve:3Step  rk
   !1!1
!
.. 1


rnr
n
Cei r
n
Proof
11
1


 r
n
r
n
r
n
CCC
r
n
r
n
r
n
CCC  

 1
1
1
 
     !!
!
!!1
!1
rnr
n
rnr
n





   
   !!1
1!!1
rnr
rnn



 
   !!1
11!
rnr
rnn




19.  
   !!1
11!
rnr
rnn




20.  
   !!1
11!
rnr
rnn



 
   !!1
!
rnr
rnn




21.  
   !!1
11!
rnr
rnn



 
   !!1
!
rnr
rnn



   !1!1
!


rnr
n

22.  
   !!1
11!
rnr
rnn



 
   !!1
!
rnr
rnn



   !1!1
!


rnr
n
Hence the result is true for k = r + 1 if it is also true for k = r

23.  
   !!1
11!
rnr
rnn



 
   !!1
!
rnr
rnn



   !1!1
!


rnr
n
Hence the result is true for k = r + 1 if it is also true for k = r
Step 4: Hence the result is true for all positive integral values of n
by induction

24. The Binomial Theorem

25. The Binomial Theorem
  n
n
nk
k
nnnnn
xCxCxCxCCx  2
2101

26. The Binomial Theorem
  n
n
nk
k
nnnnn
xCxCxCxCCx  2
2101


n
k
k
k
n
xC
0

27. The Binomial Theorem
  n
n
nk
k
nnnnn
xCxCxCxCCx  2
2101


n
k
k
k
n
xC
0
 
integerpositiveaisand
!!
!
where n
knk
n
Ck
n



28. The Binomial Theorem
  n
n
nk
k
nnnnn
xCxCxCxCCx  2
2101


n
k
k
k
n
xC
0
 
integerpositiveaisand
!!
!
where n
knk
n
Ck
n


NOTE: there are (n + 1) terms

29. The Binomial Theorem
  n
n
nk
k
nnnnn
xCxCxCxCCx  2
2101


n
k
k
k
n
xC
0
 
integerpositiveaisand
!!
!
where n
knk
n
Ck
n


NOTE: there are (n + 1) terms
This extends to;
  


n
k
kkn
k
nn
baCba
0

30. The Binomial Theorem
  n
n
nk
k
nnnnn
xCxCxCxCCx  2
2101


n
k
k
k
n
xC
0
 
integerpositiveaisand
!!
!
where n
knk
n
Ck
n


NOTE: there are (n + 1) terms
This extends to;
  


n
k
kkn
k
nn
baCba
0
4
11
Evaluate.. Cge

31. The Binomial Theorem
  n
n
nk
k
nnnnn
xCxCxCxCCx  2
2101


n
k
k
k
n
xC
0
 
integerpositiveaisand
!!
!
where n
knk
n
Ck
n


NOTE: there are (n + 1) terms
This extends to;
  


n
k
kkn
k
nn
baCba
0
4
11
Evaluate.. Cge
!7!4
!11
4
11
C

32. The Binomial Theorem
  n
n
nk
k
nnnnn
xCxCxCxCCx  2
2101


n
k
k
k
n
xC
0
 
integerpositiveaisand
!!
!
where n
knk
n
Ck
n


NOTE: there are (n + 1) terms
This extends to;
  


n
k
kkn
k
nn
baCba
0
4
11
Evaluate.. Cge
!7!4
!11
4
11
C
1234
891011




33. The Binomial Theorem
  n
n
nk
k
nnnnn
xCxCxCxCCx  2
2101


n
k
k
k
n
xC
0
 
integerpositiveaisand
!!
!
where n
knk
n
Ck
n


NOTE: there are (n + 1) terms
This extends to;
  


n
k
kkn
k
nn
baCba
0
4
11
Evaluate.. Cge
!7!4
!11
4
11
C
1234
891011




34. The Binomial Theorem
  n
n
nk
k
nnnnn
xCxCxCxCCx  2
2101


n
k
k
k
n
xC
0
 
integerpositiveaisand
!!
!
where n
knk
n
Ck
n


NOTE: there are (n + 1) terms
This extends to;
  


n
k
kkn
k
nn
baCba
0
4
11
Evaluate.. Cge
!7!4
!11
4
11
C
1234
891011



3

35. The Binomial Theorem
  n
n
nk
k
nnnnn
xCxCxCxCCx  2
2101


n
k
k
k
n
xC
0
 
integerpositiveaisand
!!
!
where n
knk
n
Ck
n


NOTE: there are (n + 1) terms
This extends to;
  


n
k
kkn
k
nn
baCba
0
4
11
Evaluate.. Cge
!7!4
!11
4
11
C
1234
891011



330
3

36. (ii) Find the value of n so that;

37. (ii) Find the value of n so that;
85a) CC nn


38. (ii) Find the value of n so that;
85a) CC nn

kn
n
k
n
CC 

39. (ii) Find the value of n so that;
85a) CC nn

kn
n
k
n
CC 
13
58


n
n

40. (ii) Find the value of n so that;
85a) CC nn

kn
n
k
n
CC 
13
58


n
n
8
20
87b) CCC nn


41. (ii) Find the value of n so that;
85a) CC nn

kn
n
k
n
CC 
13
58


n
n
8
20
87b) CCC nn

k
n
k
n
k
n
CCC  

1
11

42. (ii) Find the value of n so that;
85a) CC nn

kn
n
k
n
CC 
13
58


n
n
8
20
87b) CCC nn

k
n
k
n
k
n
CCC  

1
11
19n

43. (ii) Find the value of n so that;
85a) CC nn

kn
n
k
n
CC 
13
58


n
n
8
20
87b) CCC nn

k
n
k
n
k
n
CCC  

1
11
19n
 
11
3
5ofexpansionin the5th termtheFind 




 
b
aiii

44. (ii) Find the value of n so that;
85a) CC nn

kn
n
k
n
CC 
13
58


n
n
8
20
87b) CCC nn

k
n
k
n
k
n
CCC  

1
11
19n
 
11
3
5ofexpansionin the5th termtheFind 




 
b
aiii
 
k
k
kk
b
aCT 







3
5
1111
1

45. (ii) Find the value of n so that;
85a) CC nn

kn
n
k
n
CC 
13
58


n
n
8
20
87b) CCC nn

k
n
k
n
k
n
CCC  

1
11
19n
 
11
3
5ofexpansionin the5th termtheFind 




 
b
aiii
 
k
k
kk
b
aCT 







3
5
1111
1
 
4
7
4
11
5
3
5 





b
aCT

46. (ii) Find the value of n so that;
85a) CC nn

kn
n
k
n
CC 
13
58


n
n
8
20
87b) CCC nn

k
n
k
n
k
n
CCC  

1
11
19n
 
11
3
5ofexpansionin the5th termtheFind 




 
b
aiii
 
k
k
kk
b
aCT 







3
5
1111
1
 
4
7
4
11
5
3
5 





b
aCT
4
747
4
11
35
b
aC


47. (ii) Find the value of n so that;
85a) CC nn

kn
n
k
n
CC 
13
58


n
n
8
20
87b) CCC nn

k
n
k
n
k
n
CCC  

1
11
19n
 
11
3
5ofexpansionin the5th termtheFind 




 
b
aiii
 
k
k
kk
b
aCT 







3
5
1111
1
 
4
7
4
11
5
3
5 





b
aCT
4
747
4
11
35
b
aC
 unsimplified

48. (ii) Find the value of n so that;
85a) CC nn

kn
n
k
n
CC 
13
58


n
n
8
20
87b) CCC nn

k
n
k
n
k
n
CCC  

1
11
19n
 
11
3
5ofexpansionin the5th termtheFind 




 
b
aiii
 
k
k
kk
b
aCT 







3
5
1111
1
 
4
7
4
11
5
3
5 





b
aCT
4
747
4
11
35
b
aC
 unsimplified
4
7
8178125330
b
a


49. (ii) Find the value of n so that;
85a) CC nn

kn
n
k
n
CC 
13
58


n
n
8
20
87b) CCC nn

k
n
k
n
k
n
CCC  

1
11
19n
 
11
3
5ofexpansionin the5th termtheFind 




 
b
aiii
 
k
k
kk
b
aCT 







3
5
1111
1
 
4
7
4
11
5
3
5 





b
aCT
4
747
4
11
35
b
aC
 unsimplified
4
7
8178125330
b
a

4
7
2088281250
b
a


50.  
9
2
2
1
3inoftindependentermObtain the 




 
x
xxiv

51.  
9
2
2
1
3inoftindependentermObtain the 




 
x
xxiv
 
k
k
kk
x
xCT 







2
1
3
929
1

52.  
9
2
2
1
3inoftindependentermObtain the 




 
x
xxiv
 
k
k
kk
x
xCT 







2
1
3
929
1
0
withtermmeansoftindependenterm xx

53.  
9
2
2
1
3inoftindependentermObtain the 




 
x
xxiv
 
k
k
kk
x
xCT 







2
1
3
929
1
0
withtermmeansoftindependenterm xx
    0192
xxx
kk


54.  
9
2
2
1
3inoftindependentermObtain the 




 
x
xxiv
 
k
k
kk
x
xCT 







2
1
3
929
1
0
withtermmeansoftindependenterm xx
    0192
xxx
kk

6
0318
0218


 
k
k
xxx kk

55.  
9
2
2
1
3inoftindependentermObtain the 




 
x
xxiv
 
k
k
kk
x
xCT 







2
1
3
929
1
0
withtermmeansoftindependenterm xx
    0192
xxx
kk

6
0318
0218


 
k
k
xxx kk
 
6
32
6
9
7
2
1
3 





x
xCT

56.  
9
2
2
1
3inoftindependentermObtain the 




 
x
xxiv
 
k
k
kk
x
xCT 







2
1
3
929
1
0
withtermmeansoftindependenterm xx
    0192
xxx
kk

6
0318
0218


 
k
k
xxx kk
 
6
32
6
9
7
2
1
3 





x
xCT
6
3
6
9
2
3C


57.  
9
2
2
1
3inoftindependentermObtain the 




 
x
xxiv
 
k
k
kk
x
xCT 







2
1
3
929
1
0
withtermmeansoftindependenterm xx
    0192
xxx
kk

6
0318
0218


 
k
k
xxx kk
 
6
32
6
9
7
2
1
3 





x
xCT
6
3
6
9
2
3C

Exercise 5D; 2, 3, 5, 7, 9, 10ac, 12ac, 13,
14, 15ac, 19, 25