Math  /  Discrete

QuestionUse counting arguments from the preceding chapter.
The following eight teams will be participating in Urban University's hockey intramural tournament: the Independent Wildcats, the Phi Chi Bulldogs, the Gate Crashers, the Slide Rule Nerds, the Neural Nets, the Edmontons Eulers, the Cyber Cyborgs, and the City Slickers, Prizes will be awarded for the winner and runner-up. (a) Find the cardinality n(S)n(S) of the sample space SS of all possible outcomes of the tournament. (An outcome of the tournament consists of a winner and a runner-up.) \square (b) Let EE be the event that the City Slickers are runners-up, and let FF be the event that the Independent Wildcats are neither the winners nor runners-up. Express the event EFE \cup F in words. EFE \cup F is the event that either the City Slickers are not runners-up, and the Independent Wildcats are not the winners or runners-up. EFE \cup F is the event that either the City Slickers are not runners-up, or the Independent Wildcats are neither the winners nor runners-up. EFE \cup F is the event that either the City Slickers are runners-up, or the Independent Wildcats are neither the winners nor runners-up. EFE \cup F is the event that the City Slickers are not runners-up, and the Independent Wildcats are neither the winners nor runners-up. EFE \cup F is the event that the City Slickers are runners-up, and the Independent Wildcats are neither the winners nor runners-up.
Find cardinality n(EF)n(E \cup F). \square

Studdy Solution
Calculate n(EF) n(E \cup F) .
To find n(EF) n(E \cup F) , we use the principle of inclusion-exclusion:
n(EF)=n(E)+n(F)n(EF) n(E \cup F) = n(E) + n(F) - n(E \cap F)
- Calculate n(E) n(E) : If the City Slickers are runners-up, there are 7 choices for the winner.
n(E)=7 n(E) = 7
- Calculate n(F) n(F) : If the Independent Wildcats are neither winners nor runners-up, there are 7 choices for the winner and 6 choices for the runner-up.
n(F)=7×6=42 n(F) = 7 \times 6 = 42
- Calculate n(EF) n(E \cap F) : If the City Slickers are runners-up and the Independent Wildcats are neither winners nor runners-up, there are 6 choices for the winner.
n(EF)=6 n(E \cap F) = 6
Substitute these into the inclusion-exclusion formula:
n(EF)=7+426=43 n(E \cup F) = 7 + 42 - 6 = 43
The cardinality n(EF) n(E \cup F) is:
43 \boxed{43}

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