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Using the combination formula, it is found that he has 232 choices.
The order in which the sights are seen is not important, hence the combination formula is used to solve this question.
[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by:
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
In this problem, the choices are 8, 9, 10 or 11 sights from a set of 11, hence:
[tex]T = C_{11,8} + C_{11,9} + C_{11,10} + C_{11,11} = \frac{11!}{3!8!} + \frac{11!}{2!9!} + \frac{11!}{10!1!} + \frac{11!}{0!11!} = 232[/tex]
He has 232 choices.
You can learn more about the combination formula at https://brainly.com/question/25821700