Victor is picking out some movies to rent, and he is primarily interested in comedies and mysteries. he has narrowed down his selections to 10 comedies and 10 mysteries. how many different combinations of 3 movies can he rent if he wants at least one comedy?

1020 ways

Step-by-step explanation:

Let C= comedy

Let M= Mystery

Number of movies(comedy) = 10

Number of movies (mystery) = 10

The total number of movies from which selection is to be made is 10 + 10

= 20

We want to pick 3 movies from 10 comedies and 10 mysteries with at least one mystery.

The selection will be from;

Unconditional selection - (0 mysteries)

=unconditional selection - ( all comedy included)

= 20C3 - 10C3 For combination, [nCr = n! /(n-r) r! ]

= [20!/(20-3)!3!] - [10!/(10-3)!3!

= (6840/6) - (720/6)

= 1140 - 120

= 1020 ways

answer: .64

step-by-step explanation

set up the formula for standard deviation for this set of numbers, then simplify