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kvasek [131]
3 years ago
8

Mykel is picking out some movies to rent, and he has narrowed down his selections to 66 foreign films, 33 children's movies, 66

dramas, and 44 documentaries. How many different combinations of 99 movies can he rent if he wants all 66 dramas?
Mathematics
1 answer:
uranmaximum [27]3 years ago
7 0

Answer:

Mykel can make 2.79 x 10³² different combinations

Step-by-step explanation:

We can solve this problem using combination

The formula for combination is ⁿCₐ = \frac{n!}{(n - a)!a!}

where n = total number of choices available

a = number of items being chosen

The total movies Mykel wants is 99.

Taking out the sure 66 drama movies,

Mykel is left with 99 - 66 choices = 33 choices.

The total number of choices he is left with is 66 foreign + 33 children + 44 documentaries = 143 movies

Hence, our n = 143 movies

and a = 33 choices

Hence we do, ¹⁴⁴C₃₃ = \frac{143!}{(143 - 33)!33!} = \frac{143!}{110!33!} 2.79 x 10³² combinations

∴ Mykel can make 2.79 x 10³² different combinations.

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