Answer:
The total number of possible outcomes = 455
Step-by-step explanation:
Given:
Total number of games = 15
Number of games to be chosen = 3
The total number of ways in which the 3 games out of 15 be choosen can be given by:
[tex]_nC_r=\frac{n!}{r!(n-r)!}[/tex]
where [tex]n\rightarrow[/tex] represents total number of games and
[tex]r\rightarrow[/tex] represents the number of choices to be made.
Plugging in the known values:
[tex]15C3[/tex]
⇒ [tex]\frac{15!}{3!(15-3)!}[/tex]
⇒ [tex]\frac{15!}{3!(12)!}[/tex]
⇒ [tex]\frac{15\times14\times13\times12!}{3!\times12!}[/tex] [Canceling out the common terms]
⇒ [tex]\frac{15\times14\times13}{3\times2\times1}[/tex]
⇒ [tex]455[/tex]