Question

part 1 was solved in picture. Lee Ann is picking out some movies to rent, and she is primarily interested in children's movies and comedies. She has narrowed down her selection to 18 children's movies and 7 colonies.part 2 : how many different combinations of 3 movies can she rent if she once at least one comedy?

Answer 1

Step-by-step explanation:

Total movies = children movies + comedies = 18 + 7 = 25

The combinations of 3 movies that she can rent if she once at least one comedy:

`\begin{gathered} combination(1\text{ comedy \$ 2 ch i}ldren\text{ movies) + }combination(2\text{ comedy \$ 1 ch i}ldren\text{ movies) } \\ +\text{ }combination(3\text{ comedy \$ 0 ch i}ldren\text{ movies) } \end{gathered}`
`\begin{gathered} ^{}\text{combination formula: } \\ ^nC_r\text{ = }\frac{n}{(n\text{ - r)!r!}} \\ combination(1\text{ comedy \$ 2 ch i}ldren\text{ movies) = }^7C_1*^(18)C_2 \\ =\text{ }\frac{7!}{(7\text{- 1)!1!}}*\frac{18!}{(18\text{ - 2)!2!}}=\text{ 7}*153 \\ =\text{ 1071} \end{gathered}`
`\begin{gathered} combination(2\text{ comedy \$ 1 ch i}ldren\text{ movies) = }^7C_2*^(18)C_1 \\ =\text{ }\frac{7!}{(7\text{- 2)!2!}}*\frac{18!}{(18\text{ - 1)!1!}}=\text{ 21}*18 \\ =378 \end{gathered}`
`\begin{gathered} combination(3\text{ comedy \$ 0 ch i}ldren\text{ movies) = }^7C_3*^(18)C_0 \\ =\text{ }\frac{7!}{(7\text{- 3)!3!}}*\frac{18!}{(18\text{ - 0)!0!}}=\text{ 35}*1 \\ =35 \end{gathered}`
`\begin{gathered} The\text{ }combinations\text{ }of\text{ }3\text{ }movies\text{ }that\text{ }she\text{ }can\text{ }rent\text{ if }shewantsatleastonecomedy\text{ } \\ =\text{ 1071 + 378 + 35} \\ =\text{ 1484} \end{gathered}`