1. Mathematics for Computer Science MIT 6.042J/18.062J
Combinatorics II
Copyright © Radhika Nagpal, 2002.
Prof. Albert Meyer & Dr. Radhika Nagpal

2. Last Week: Counting I Sets Pigeonhole Principle Permutations
Bijections, Sum Rule, Inclusion-Exclusion, Product Rule
Pigeonhole Principle
Permutations
Tree Diagrams

3. This Week: Counting II Division Rule
Combinations (binomial coefficients)
Binomial Theorem and Identities
Permutations with limited repetition (multinomial coefficients)
Combinations with repetition (stars and bars)

4. The Real Agenda: Poker
How many different hands could I get, if I played 5-card draw?

5. The Real Agenda: Poker (52) (51) (50) (49) (48)
How many different hands could I get, if I played 5-card draw?
(52) (51) (50) (49) (48)

6. The Real Agenda: Poker (52) (51) (50) (49) (48)
How many different hands could I get, if I played 5-card draw?
(52) (51) (50) (49) (48)
r-permutation, P(52,5)

7. Problem: Over-counting
These two hands are the same:

8. Problem: Over-counting
These two hands are the same:

9. Problem: Over-counting
These two hands are the same:
In fact any permutation of these cards is the same hand (order is irrelevant)

10. Number of 5-card Hands
How much have we over-counted?

11. Number of 5-card Hands How much have we over-counted?
Over-counted EVERY HAND by 5!

12. Number of 5-card Hands …… How much have we over-counted?
Over-counted EVERY HAND by 5! ……

13. Number of 5-card Hands …… How much have we over-counted?
Over-counted EVERY HAND by 5! ……
Still approximately 2.5 million possible hands

14. Combinations
C(n,r) = number of different subsets of size r from a set with n elements.
C(n,r) = P(n,r) / r!

15. Combinations
C(n,r) = P(n,r)/r! =

16. Combinations
C(n,r) = P(n,r)/r! =

17. Combinations
C(n,r) = P(n,r)/r! =

18. Combinations
C(n,r) = P(n,r)/r! =

19. Poker: Gambling Table Straight Flush > Four-of-a-kind
> Full House
> Flush
> Straight
> Three-of-a-kind
> Two pair
> One pair
> No pair

20. Poker: Four-of-a-kind
Card: value (13) + suit (4)
Four-of-a-kind
4 cards with the same value
1 with a different value
EXAMPLE: 9s 9d 9c 9h 5h

21. Poker: Four-of-a-kind
2 spades
2 clubs
2 hearts
2 diamonds .
K hearts
K diamonds 1 2 3 .
11 (jack)
12 (queen)
13 (king)

22. Poker: Four-of-a-kind
2 spades
2 clubs
2 hearts
2 diamonds .
K hearts
K diamonds 1 2 3 .
11 (jack)
12 (queen)
13 (king)
Four-of-a-kind = = 624

23. Poker: Full House Full House 9s 9c 9d 5s 5c
= choose value for the triple + choose 3 suits + choose value for pair + choose 2 suits

24. In-class Problem 1: Each table should write their solution on their whiteboard.

25. Incorrect Counting Argument
Two pair =
Every hand is counted twice

26. Correct Counting Arguments
1. Divide the Theory Pig’s estimate by 2
2. Choose the values of the two pairs together
2-pair =

27. Division Rule If set B over-counts every element of A by k times then,
|B| = k |A|

28. Permutations vs Combinations
Combinations: subsets of size r,
order does not matter
Permutations: strings of length r,
order of elements does matter.

29. Calculating Permutations and Combinations
Closely related: C(n,r) = P(n,r) / r!
C(n,r) = count all r-permutations, every combination is over-counted by r!
P(n,r) = choose r items, then take all permutations of the items

30. Counting Powerset of A P(A) = set of all subsets of A A = {a,b}
P(A) = {{}, {a}, {b}, {a,b}}

31. Counting P(A) Bijection : P(A) and binary strings of length |A|
A = {a1 a2 a3 a4 a5……an}
Binary String = …..0
Subset of A = {a1, a3, a5}
| P(A) | = 2n

32. Counting P(A) |P(A)| = = subsets + subsets + subsets…..subsets
of size 0 of size 1 of size 2 of size n

33. Counting P(A) Identity: |P(A)| =
= subsets + subsets + subsets…..subsets
of size 0 of size 1 of size 2 of size n
Identity:

34. Poker: Gambling Table Straight Flush = 40 Four-of-a-kind = 624
Full house = 3744
Flush = 5148
Straight = 10240
Three-of-a-kind = 54,912
Two pair =123,552
One pair = 1,098,240
No pair = 1,317,388

35. In-class Problems