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Binomial Coefficient
Supplementary Notes
Prepared by Raymond WongPresented by Raymond Wong
2
e.g.1 (Page 4) Prove that
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0
24
i i
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i i
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e.g.1A set of all possible subsets of {1, 2, 3, 4}
{}
{1}
{2}
{3}
{4}
{1, 2}
{1, 4}
{2, 4}
{1, 3}
{2, 3}
{3, 4}
{1, 2, 3}
{1, 2, 4}
{1, 3, 4}
{2, 3, 4}
{1, 2, 3, 4}
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0
24
i i
{}
{1}
{2}
{3}
{4}
{1, 2}
{1, 4}
{2, 4}
{1, 3}
{2, 3}
{3, 4}
{1, 2, 3}
{1, 2, 4}
{1, 3, 4}
{2, 3, 4}
{1, 2, 3, 4}
S0
S1 S2 S3
S4
4
e.g.1A set of all possible subsets of {1, 2, 3, 4}
{}
{1}
{2}
{3}
{4}
{1, 2}
{1, 4}
{2, 4}
{1, 3}
{2, 3}
{3, 4}
{1, 2, 3}
{1, 2, 4}
{1, 3, 4}
{2, 3, 4}
{1, 2, 3, 4}
{}
{1}
{2}
{3}
{4}
{1, 2}
{1, 4}
{2, 4}
{1, 3}
{2, 3}
{3, 4}
{1, 2, 3}
{1, 2, 4}
{1, 3, 4}
{2, 3, 4}
{1, 2, 3, 4}
S0
S1 S2 S3
S4
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41
42
43
44
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24
i i
+ + + +
5
e.g.1A set of all possible subsets of {1, 2, 3, 4}
{}
{1}
{2}
{3}
{4}
{1, 2}
{1, 4}
{2, 4}
{1, 3}
{2, 3}
{3, 4}
{1, 2, 3}
{1, 2, 4}
{1, 3, 4}
{2, 3, 4}
{1, 2, 3, 4}
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i i
{2}
1 does not appear2 appears3 does not appear4 does not appear
{2, 3, 4}
1 does not appear2 appears3 appears4 appears
We can have another representation (related to “whether an element appears or not”)to represent a subset
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e.g.1A set of all possible subsets of {1, 2, 3, 4}
{}
{1}
{2}
{3}
{4}
{1, 2}
{1, 4}
{2, 4}
{1, 3}
{2, 3}
{3, 4}
{1, 2, 3}
{1, 2, 4}
{1, 3, 4}
{2, 3, 4}
{1, 2, 3, 4}
44
0
24
i i
1
Appears
Does notappear
2
Appears
Does notappear
3
Appears
Does notappear
4
Appears
Does notappear
{2}
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e.g.1A set of all possible subsets of {1, 2, 3, 4}
{}
{1}
{2}
{3}
{4}
{1, 2}
{1, 4}
{2, 4}
{1, 3}
{2, 3}
{3, 4}
{1, 2, 3}
{1, 2, 4}
{1, 3, 4}
{2, 3, 4}
{1, 2, 3, 4}
44
0
24
i i
1
Appears
Does notappear
2
Appears
Does notappear
3
Appears
Does notappear
4
Appears
Does notappear
{2, 3, 4}
8
e.g.1A set of all possible subsets of {1, 2, 3, 4}
{}
{1}
{2}
{3}
{4}
{1, 2}
{1, 4}
{2, 4}
{1, 3}
{2, 3}
{3, 4}
{1, 2, 3}
{1, 2, 4}
{1, 3, 4}
{2, 3, 4}
{1, 2, 3, 4}
44
0
24
i i
1
Appears
Does notappear
2
Appears
Does notappear
3
Appears
Does notappear
4
Appears
Does notappear
2 choices 2 choices 2 choices 2 choices
Total number of subsets of {1, 2, 3, 4} = 2 x 2 x 2 x 2 = 24
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e.g.2 (Page 16) Prove that
2
4
1
4
2
5
2
4
1
4
2
5
10
e.g.2
2
4
1
4
2
5
S
A
B
C
D
E
{A, B}
{A, C}
{A, D}
{B, C}
{B, D}
{C, D}
{A, E}
{B, E}
{C, E}
{D, E}
A set of all possible 2-subsets of S
52
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e.g.2
2
4
1
4
2
5
S
A
B
C
D
E
{A, B}
{A, C}
{A, D}
{B, C}
{B, D}
{C, D}
{A, E}
{B, E}
{C, E}
{D, E}
A set of all possible 2-subsets of S
{A, B}
{A, C}
{A, D}
{B, C}
{B, D}
{C, D}
A set of all possible 2-subsets of S notcontaining E
{A, E}
{B, E}
{C, E}
{D, E}
A set of all possible 2-subsets of S containing E
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e.g.2
2
4
1
4
2
5
S
A
B
C
D
E
{A, B}
{A, C}
{A, D}
{B, C}
{B, D}
{C, D}
A set of all possible 2-subsets of S notcontaining E
{A, E}
{B, E}
{C, E}
{D, E}
A set of all possible 2-subsets of S containing E
A set of all possible 2-subsets of {A, B, C, D} S’
A
B
C
D
42
We know that each 2-subset contains E.Since each 2-subset contains 2 elements, the other ONE element comes from {A, B, C, D} S’
A
B
C
D
41
+
This proof is an example of a combinatorial proof.
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e.g.3 (Page 22) Prove that
32233
3
3
2
3
1
3
0
3)( yxyyxxyx
32233
3
3
2
3
1
3
0
3)( yxyyxxyx
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e.g.3
32233
3
3
2
3
1
3
0
3)( yxyyxxyx
x
y
blue
x
y
red
x
y
green
monomial
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e.g.3
32233
3
3
2
3
1
3
0
3)( yxyyxxyx
y
Set of y’s (in different colors)
y
y
Coefficient of xy2=No. of ways of choosing 2 y’s from this set32=
Two y’s are in different colors.
Interpretation 1
Suppose that we choose 2 elements.These two elements correspond to two y’s indifferent colors.
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e.g.3
32233
3
3
2
3
1
3
0
3)( yxyyxxyx
Coefficient of xy2=No. of lists containing 2 y’s 32=
L1 L2 L3 Each Li can be x or y.Now, I want to have 2 y’s in this list. 1
Set of positions in the list
2
3Suppose that we choose 2 positions from the set.These two positions correspond to the positions thaty appears.
e.g., {1, 3} means yxy
Each monomial has 3 elements.Each element can be x or y.
Interpretation 2
These 2 y’s appear in 2 different positions in this list.
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e.g.3
32233
3
3
2
3
1
3
0
3)( yxyyxxyx
Coefficient of xy2=
L1 L2 L3
1
Set of positions in the list
2
3
Interpretation 3
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e.g.3
32233
3
3
2
3
1
3
0
3)( yxyyxxyx
Coefficient of xy2=No. of ways of distributing 3 objects 32=
L1 L2 L3
1
Set of positions in the list
2
3 Choose 2 objects
Bucket B132
Bucket B2
Interpretation 3
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e.g.4 (Page 29)
Suppose we have 2 distinguishable buckets, namely B1 and B2
How many ways can we distribute 5 objects into these buckets such that 2 objects are in B1
3 objects are in B2?
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e.g.4
Choose 2 objects
B1
B2
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e.g.5 (Page 30)
Suppose we have 3 distinguishable buckets, namely B1, B2 and B3.
How many ways can we distribute 9 objects into these buckets such that 2 objects are in B1
3 objects are in B2, and 4 objects are in B3?
9 objects 2 objects in B1, 3 objects in B2 and 4 objects in B3
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e.g.5
Choose 2 objects
B1
92
9 objects 2 objects in B1, 3 objects in B2 and 4 objects in B3
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e.g.5
Choose 2 objects
B1
92
Choose 3 objects
B273
B3
Total number of placing objects = 92
73
x9!
2!(9-2)! 3!(7-3)!
7!x=
9!
2!3!4!=
9 objects 2 objects in B1, 3 objects in B2 and 4 objects in B3
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e.g.6 (Page 33) Prove that
the coefficient of in (x + y + z)4
is equal to
where k1 + k2 + k3 = 4
321 kkk zyx
321
4
kkk
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e.g.6
Coefficient of x2yz= 42 1 1
=
L1 L2 L3
L4
Choose 2 objects
Bucket B1
Bucket B2
(x + y + z) (x + y + z) (x + y + z) (x + y + z)= xxxx + xxxy + xxxz + xxyx + … + zzzy + zzzz
1
Set of positions in the list
2
3
4
Choose 1 object Bucket B3
