1. 1 Binomial Coefficient Supplementary Notes Prepared by Raymond Wong Presented by Raymond Wong

2. 2 e.g.1 (Page 4) Prove that

3. 3 e.g.1 A 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} S0S0 S1S1 S2S2 S3S3 S4S4

4. 4 e.g.1 A 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} S0S0 S1S1 S2S2 S3S3 S4S4 4040 4141 4242 4343 4444 + + ++

5. 5 e.g.1 A 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} {2} 1 does not appear 2 appears 3 does not appear 4 does not appear {2, 3, 4} 1 does not appear 2 appears 3 appears 4 appears We can have another representation (related to “whether an element appears or not”) to represent a subset

6. 6 e.g.1 A 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 Appears Does not appear 2 Appears Does not appear 3 Appears Does not appear 4 Appears Does not appear {2}

7. 7 e.g.1 A 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 Appears Does not appear 2 Appears Does not appear 3 Appears Does not appear 4 Appears Does not appear {2, 3, 4}

8. 8 e.g.1 A 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 Appears Does not appear 2 Appears Does not appear 3 Appears Does not appear 4 Appears Does not appear 2 choices Total number of subsets of {1, 2, 3, 4} = 2 x 2 x 2 x 2 = 2 4

9. 9 e.g.2 (Page 16) Prove that

10. 10 e.g.2 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 5252

11. 11 e.g.2 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 not containing E {A, E} {B, E} {C, E} {D, E} A set of all possible 2-subsets of S containing E

12. 12 e.g.2 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 not containing 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 4242 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 4141 + This proof is an example of a combinatorial proof.

13. 13 e.g.3 (Page 22) Prove that

14. 14 e.g.3 x y blue x y red x y green monomial

15. 15 e.g.3 y Set of y’s (in different colors) y y Coefficient of xy 2 =No. of ways of choosing 2 y’s from this set 3232 = Two y’s are in different colors. Interpretation 1 Suppose that we choose 2 elements. These two elements correspond to two y’s in different colors.

16. 16 e.g.3 Coefficient of xy 2 =No. of lists containing 2 y’s 3232 = L1 L2 L3L1 L2 L3 Each L i can be x or y. Now, I want to have 2 y’s in this list. 1 Set of positions in the list 2 3 Suppose that we choose 2 positions from the set. These two positions correspond to the positions that y 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.

17. 17 e.g.3 Coefficient of xy 2 = L1 L2 L3L1 L2 L3 1 Set of positions in the list 2 3 Interpretation 3

18. 18 e.g.3 Coefficient of xy 2 =No. of ways of distributing 3 objects 3232 = L1 L2 L3L1 L2 L3 1 Set of positions in the list 2 3 Choose 2 objects Bucket B 1 3232 Bucket B 2 Interpretation 3

19. 19 e.g.4 (Page 29) Suppose we have 2 distinguishable buckets, namely B 1 and B 2 How many ways can we distribute 5 objects into these buckets such that 2 objects are in B 1 3 objects are in B 2 ?

20. 20 e.g.4 Choose 2 objects B1B1 B2B2 5252

21. 21 e.g.5 (Page 30) Suppose we have 3 distinguishable buckets, namely B 1, B 2 and B 3. How many ways can we distribute 9 objects into these buckets such that 2 objects are in B 1 3 objects are in B 2, and 4 objects are in B 3 ? 9 objects  2 objects in B 1, 3 objects in B 2 and 4 objects in B 3

22. 22 e.g.5 Choose 2 objects B1B1 9292 9 objects  2 objects in B 1, 3 objects in B 2 and 4 objects in B 3

23. 23 e.g.5 Choose 2 objects B1B1 9292 Choose 3 objects B2B2 7373 B3B3 Total number of placing objects = 9292 7373 x 9! 2!(9-2)! 3!(7-3)! 7! x = 9! 2!3!4! = 9 objects  2 objects in B 1, 3 objects in B 2 and 4 objects in B 3

24. 24 e.g.6 (Page 33) Prove that the coefficient of in (x + y + z) 4 is equal to where k 1 + k 2 + k 3 = 4

25. 25 e.g.6 Coefficient of x 2 yz= 4 2 1 1 = L 1 L 2 L 3 L 4 Choose 2 objects Bucket B 1 Bucket B 2 (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 B 3