Problem 7

How many ways are there to deal a hand of five cards to two players from a deck of cards (52)?

The order of the five cards doesn’t matter, so this is a combinations problem and not a permutation problem. By the Rule of Product we multiply the number of ways to deal a hand to each player together. The first player chooses 5 cards from the 52 in the deck. The second player chooses 5 cards from the remaining 47 in the deck.

C(52, 5) \cdot C(47, 5)

\frac{52!}{\cancel{47!} \cdot5!} \cdot \frac{\cancel{47!}}{42! \cdot5!} = 3,986,646,103,440

Problem 8

A group of friends composed of Henry, Breonn, Jodi, Rich, Brandon, Scurvy, and Kat need to select four players for a video game at a party. How many ways are there to select the four players if Henry and Kat, or both, are included?

Let P_{1} represent the set of selections where Henry is included and let P_{2} represent the set of selections where Kat is included. Since P_{1} and P_{2} are not disjoint sets (both Henry and Kat could be in the same set), we can’t use the Rule of Sum.

First we count the number of ways Henry can be included, set P_{1}. He can be assigned a spot in four ways from the four open video game spots. The second spot can be chosen from the remaining six people. The third spot can be chosen from the remaining five people, and the fourth spot can be chosen from the remaining four people. So, by the Rule of Product, the number of ways Henry can be selected into a spot among the four players is

P_{1} = 4 \cdot 6 \cdot 5 \cdot 4 = 480

This same selection process applies to Kat.

P_{2} = 480

Now we have to compute the number of ways Henry and Kat are both included to play the game. Henry can be placed into any of the four positions and Kat can be placed into any of the remaining three spots. The third spot can be selected from the remaining five players, and the fourth spot can be chosen from the remaining four people. This comes out to be

|P_{1} \cap P_{2}| = 4 \cdot 3 \cdot 5 \cdot 4 = 240

By the Inclusion-Exclusion Principle

|P_{1} \cup P_{2}| = |P_{1}| + |P_{2}| - |P_{1} \cap P_{2}| = 480 + 480 - 240 = 720

There are 720 ways that Henry, Kat, or both can be selected.