# Problem: Alternative

In the reading, you learned that the number of \(k\)-sized (unordered) sets drawn from \(n\) objects is given by:

\[ {n \choose k} = \frac{(n)_k}{k!}. \]

If you have seen the combination (or “choose”) operator before, you may not have learned that it was equal to this quantity. Instead, you likely learned that:

\[ {n \choose k} = \frac{n!}{k!(n-k)!}. \]

In a few sentences, justify or “explain” this second formula in terms of overcounting. What are you overcounting and what are you removing from that overcount to arrive at the desired final quantity?

# Problem: Poker Hands

In variations of *poker*, players receive a five-card hand
from a deck of 52 player cards. Recall that each of the 52 cards has a
*rank* and *suit*:

- The ranks range from 2–10, Jack (J), Queen (Q), King (K), and Ace
(A),
*13 ranks*overall. - The suits are drawn from spades, clubs, hearts, and diamonds.

Give combinatorial descriptions for each of the following values.

The total number of possible poker hands. Remember that a poker hand is an

*unordered collection of five cards*.The total number of possible poker hand that contains

*exactly*two pairs. A pair is a pair of the cards with the same rank,*e.g.*, the 8 of spades and hearts.Rather than computing this quantity by determining the ways of drawing the first card of the hand, the second card, and so forth, which will lead to a series of conditional choices, consider the following strategy instead:

- Choose 2 ranks to participate in the pair.
- Choose 2 suits from each rank to participate in the pair.
- Choose a remaining rank and a suite from that rank as the final card.

The total number of possible poker hands that are a

*full house*. A full house is a hand consisting of a three-of-a-kind and a pair. A three-of-a-kind is a pair but with three cards of the same rank instead of two.The total number of possible poker hands that are a four-of-a-kind,

*e.g.*, the Ace of spades, clubs, hearts, and diamonds.The total number of possible poker hands that are a

*single pair*. A single pair is where two of the cards have the same rank,*e.g.*, the Jack of spades and hearts. Note that when a hand contains a single pair, it*should not*contain other, better hands,*e.g.*, two pairs, a three-of-a-kind, or a four-of-a-kind.The total number of possible poker hands that are a

*flush*where all five cards are of the same suit,*e.g.*, all spades cards. Like a pair, this quantity should also not include better hands: a straight flush (where you have a flush*and*the cards are in consecutive rank,*e.g.*, 5-6-7-8-9) and a royal flush (a straight flush that starts with 10,*i.e.*, 10-J-Q-K-A).

Check your work by observing that your combinatorial descriptions produce these quantities:

- The number of poker hands: 2,598,960.
- Two pairs: 123,552.
- Full house: 3,744.
- Four-of-a-kind: 624.
- Pair: 1,098,240.
- Flush: 5,108.

# Problem: Deceptive

With five-card poker hands, a royal flush is a straight flush that runs 10-J-Q-K-A. The following combinatorial description denotes the number of royal flushes:

\[ {4 \choose 1}. \]

In a sentence or two, justify this combinatorial description.

# Problem: Shades of Pre-registration

Suppose that you are building a schedule from among \(k\) distinct possible courses. Give combinatorial descriptions of each of the cardinalities described below. For each description, check your work by instantiating a concrete set of courses, and demonstrating that your description works for each such concrete example.

The number of

*five course schedules*(*i.e.*, sequences of courses) you can build from these courses.The number of

*sets of three courses*(*i.e.*, unordered collections) you can build from these courses.The number of

*four course schedules*that contain both Basket Weaving or Underwater Knitting.(

*Hint*: choose positions for Basket Weaving and Underwater Knitting. This implies where the remaining two courses will go. Finally choose which courses go into those remaining two slots.)The number of

*four course schedules*that*do not contain*both Basket Weaving and Underwater Knitting.(

*Hint*: leverage your answer to the previous part!)The number of

*six course schedules*that include Calculus I and Computer Science I but do not place Calculus I*after*Computer Science I in the schedule.(

*Hint*: similarly to the previous part, choose positions for the two named courses. But here, you need to remove the possibilities that get the order these courses swapped.)(

**Bonus Problem**) The number of*six course schedules*that include Calculus I and Computer Science I but do not place Calculus I*immediately after*Computer Science I in the schedule.(

*Hint*: unlike the previous problem, we only want to ensure that Calc I does not appear in the next time slot after CS I. How can we break up the possible positions of the two courses to get a handle on these situations? How do we then account for overcounting?)