CT 2 Discrete Math CSE 103 resources.

Date: May 23, 2025

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My own Notes: 

CT and Term's QnA:

Topic analysis with chatgpt for better output:

Important basics:

Slide 4: Proofs by contrapositive ***, Direct proof*, Proof  by cases**,

Basic Counting

Topic : poker hand:

What is a poker hand?

• A poker hand consists of 5 cards drawn from a standard deck of 52 playing cards.

• Each card is unique.

• We want to count how many different poker hands are possible.

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Important:

• The order of cards does NOT matter in a hand. (A hand with cards A, K, Q, J, 10 is the same no matter the order you hold them.)

• So this is a combination problem, not a permutation.

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Step 1: Total number of ways to choose 5 cards out of 52

Using combinations:

$$\text{Number of poker hands} = \binom{52}{5} = \frac{52!}{5! \times (52 - 5)!}$$

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Step 2: What does $\binom{52}{5}$ mean?

• Choose any 5 cards out of 52 without caring about the order.

• This counts all possible 5-card hands.

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Step 3: Calculate the number (approximate)

$$\binom{52}{5} = \frac{52 \times 51 \times 50 \times 49 \times 48}{5 \times 4 \times 3 \times 2 \times 1} = 2,598,960$$

So there are 2,598,960 possible poker hands.

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Step 4: Counting specific types of hands (example)

Say you want to count how many hands have exactly 2 Aces.

• Number of Aces in deck: 4

• Choose 2 Aces out of 4: $\binom{4}{2} = 6$

• Remaining 3 cards must be chosen from the 48 non-Ace cards: $\binom{48}{3} = 17,296$

Total such hands:

$$6 \times 17,296 = 103,776$$

// 2 Aces are similar. so it has the combo of 6; then  I have 48 cards in option. cause, i have exactly only 2 aces, not more than that.. so i can take 3 cards from resting 48 cards. thats the equation. there can be only 4 aces in each suit, so rest of the cards (48) are non-ace card

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Summary:

• Total poker hands = combinations of 52 cards taken 5 at a time.

• Use combination formula $\binom{n}{k}$ when order does not matter.

• To count specific types of hands, break the problem into stages and multiply counts (product rule).

Example 1: Four of a Kind A Four-of-a-Kind is a set of four cards with the same value. How many different hands contain a Four-of-a-Kind? One way to do this is to first map the problem into a problem of counting sequences.

ChatGPT said:

Great! Let’s solve the Four-of-a-Kind counting problem step by step using counting principles.

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What is Four-of-a-Kind?

• A hand that has 4 cards of the same rank (e.g., four 7s or four Kings).

• The 5th card can be any card different in rank from the four identical cards.

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Step 1: Count the choices for the rank of the Four-of-a-Kind

• There are 13 ranks in a deck (2, 3, 4, ..., 10, J, Q, K, A).

• Choose which rank you want 4 of:

  $$\binom{13}{1} = 13$$

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Step 2: For that chosen rank, choose all 4 suits

• Since you want all four cards of that rank, you must take all 4 suits.

• So only 1 way here (all 4 suits).

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Step 3: Choose the 5th card

• The 5th card cannot be the same rank as the four you already chose.

• There are $52 - 4 = 48$ cards left (because 4 cards of the chosen rank are taken).

• So, you can pick any of those 48 cards for the 5th card.

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Step 4: Multiply choices (Product Rule)

$$\text{Total Four-of-a-Kind hands} = 13 \times 1 \times 48 = 624$$

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Intuition:

• First pick the rank of the Four-of-a-Kind (13 ways).

• You have no choice about suits for the four cards (all four must be there).

• Choose any card for the 5th position (48 options).

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Mapping to sequences:

If you think of the hand as a sequence:

• First 4 cards: fixed as 4 suits of the chosen rank.

• Fifth card: any card except the rank chosen for four.

Counting the number of such sequences and realizing order doesn’t matter, you get the same count $624$.

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If you want, I can explain other poker hands (like Full House, Flush, Straight) or how the counting of sequences relates to combinations. Want me to?

You said:

A Full House is a hand with three cards of one value and two cards of another value. How many different hands contain a Full House?

ChatGPT said:

Great! Let's count the number of Full House poker hands step-by-step.

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What is a Full House?

• Three cards of the same rank (called "three-of-a-kind").

• Two cards of a different rank (called "a pair").

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Step 1: Choose the rank for the three-of-a-kind

• There are 13 ranks.

• Choose 1 rank for the three-of-a-kind:

  $$\binom{13}{1} = 13$$

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Step 2: Choose 3 suits out of 4 for that rank

• For the chosen rank, select 3 cards (suits) out of 4:

  $$\binom{4}{3} = 4$$

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Step 3: Choose the rank for the pair

• The pair must have a different rank than the three-of-a-kind.

• Remaining ranks: $13 - 1 = 12$

• Choose 1 rank for the pair:

  $$\binom{12}{1} = 12$$

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Step 4: Choose 2 suits out of 4 for the pair

• For the chosen rank of the pair, select 2 suits out of 4:

  $$\binom{4}{2} = 6$$

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Step 5: Multiply all choices (Product Rule)

$$\text{Total Full House hands} = 13 \times 4 \times 12 \times 6 = 3744$$

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Summary:

• 13 ways to choose the three-of-a-kind rank.

• 4 ways to choose 3 suits for it.

• 12 ways to choose the pair’s rank (different from three-of-a-kind).

• 6 ways to choose 2 suits for the pair.

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So, the number of Full House poker hands is:

$$\boxed{3744}$$

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If you want, I can explain other poker hand counts or help you understand the reasoning behind these combination choices!

You said:

Example 4: Every Suit How many hands contain at least one card from every suit? 1. The value of each suit, which can be selected in 13x13x13x13 ways. 2. The suit of the extra card, which can be selected in 4 ways. 3. The value of the extra card, which can be selected in 12 ways.

ChatGPT said:

You said:

Example 4: Every Suit How many hands contain at least one card from every suit? 1. The value of each suit, which can be selected in 13x13x13x13 ways. 2. The suit of the extra card, which can be selected in 4 ways. 3. The value of the extra card, which can be selected in 12 ways. Double count! So the answer is 134x4x12/2 = 685464

ChatGPT said:

Ah, you’ve pointed out a very important detail: double counting happens here! Let's clarify everything carefully and explain why the double counting occurs and how the division by 2 fixes it.

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Problem recap:

• We want to count the number of 5-card hands that contain at least one card from every suit.

• That means the hand must have cards from all four suits.

• Since the hand has 5 cards and 4 suits, one suit must appear twice (the "extra" card).

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Initial approach:

1. Choose one card from each suit:

$$13 \times 13 \times 13 \times 13 = 13^4 = 28,561$$

2. Choose the suit for the extra card:

$$4 \text{ ways}$$

3. Choose the value (rank) of the extra card in that suit:

• Since we already chose one card of that suit, we cannot choose the same card again.

• So, $13 - 1 = 12$ ways.

4. Multiply:

$$13^4 \times 4 \times 12 = 28,561 \times 4 \times 12 = 1,371,000 + \text{(approx)}$$

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The double counting issue:

Why double counting happens?

• When we count like this, we count each hand twice.

• Because the order in which we pick the "extra card" suit and the single card from that suit matters.

• For example, suppose the suit with the extra card is Hearts.

• But the way the formula is constructed, if the extra card is card A from Hearts, and the single card from Hearts is card B, then switching the roles of A and B also counts the same hand again.

• Hence, each hand with 2 cards from the same suit is counted twice.

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How to fix double counting?

• Since every such hand is counted exactly twice, divide by 2 to correct.

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Final corrected formula:

$$\frac{13^4 \times 4 \times 12}{2} = 685,464$$

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Summary:

• Initial count: $13^4 \times 4 \times 12$ (overcounts by a factor of 2).

• Corrected count: divide by 2.

• Result: $685,464$ hands contain at least one card from every suit.

Induction:

What is Mathematical Induction?

Mathematical induction is a method to prove that a statement $P(n)$ is true for all integers $n \geq k$ (usually $k = 1$ or 0).

It works kind of like a domino effect:

1. You prove the first domino falls (base case).

2. You prove that if any domino $n$ falls, then the next domino $n+1$ also falls (inductive step).

If both are true, then all the dominos fall, meaning the statement is true for all $n$.

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Structure of Proof by Mathematical Induction

Suppose we want to prove $P(n)$ is true for all $n \ge k$:

1. Base Case:
   Verify $P(k)$ is true.

2. Inductive Hypothesis:
   Assume $P(m)$ is true for some arbitrary $m \ge k$. (We don't prove it here; just assume.)

3. Inductive Step:
   Using the assumption $P(m)$, prove that $P(m+1)$ is true.

If these steps succeed, then $P(n)$ is true for all $n \ge k$.

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Simple Example

Prove:
$1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2} \text{ for all } n \ge 1$

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Step 1: Base Case $n=1$

Left side: $1$
Right side: $\frac{1(1+1)}{2} = \frac{2}{2} = 1$

Both sides equal, so base case holds.

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Step 2: Inductive Hypothesis

Assume true for $n = m$:

$1 + 2 + \cdots + m = \frac{m(m+1)}{2}$

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Step 3: Inductive Step (prove for $n = m+1$)

Add $m+1$ to both sides of the inductive hypothesis:

$$1 + 2 + \cdots + m + (m+1) = \frac{m(m+1)}{2} + (m+1)$$

Simplify the right side:

$$\frac{m(m+1)}{2} + (m+1) = \frac{m(m+1) + 2(m+1)}{2} = \frac{(m+1)(m + 2)}{2} = \frac{(m+1)((m+1)+1)}{2}$$

This matches the formula with $n = m+1$.

Thus, $P(m+1)$ is true.

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