leeye84 wrote:

Carly has 3 movies that she can watch during the weekend: 1 Action movie, 1 Comedy, and 1 Drama. However, she needs to watch the Drama 3 times. Assuming Carly has time for 5 movies and intends to watch all of them, in how many ways can she do so?

Below is a revised version of this question:

Carly has three movies that she can watch during the weekend: an action movie, a comedy, or a drama. However, she wants to watch the same drama movie three times, an action movie once and a comedy movie also once. In how many different ways can she arrange these five screenings?

A. 6

B. 20

C. 24

D. 60

E. 120

The number of different ways Carly can watch Drama, Drama, Drama, Action, Comedy (DDDAC) is basically the number of arrangements of 5 letters DDDAC out of which 3 D's are identical, so it's \frac{5!}{3!}=20.

Answer: B.

This is the way the question should be phrased to avoid ambiguity.

(3D1C1A )5!/3!+ (3D2C+3D2A)2*5!/3!2! +(4D1A+4D1C)2*5!/4! +(5D)5!/5! = 46