Question

1. an all - night showing at a movie theatre is to consist of five movies. there are fourteen different movies available, ten disaster movies and four horror movies. how many possible schedules include:

a) at least one horror movie?
b) at least four disaster movies?
c) both “airport disaster” and “halloween horror”?
permutations and combinations lesson #5: combinations - part two

Explanation:

Part a)

Step 1: Total without horror

To find the number of schedules with at least one horror movie, we first find the total number of schedules without any horror movies and subtract it from the total number of schedules. The total number of ways to choose 5 movies from all \(14 + 4=18\) movies is \(\binom{18}{5}\). The number of ways to choose 5 movies from only the 14 non - horror (disaster) movies is \(\binom{14}{5}\).
The formula for combinations is \(\binom{n}{k}=\frac{n!}{k!(n - k)!}\)
\(\binom{18}{5}=\frac{18!}{5!(18 - 5)!}=\frac{18\times17\times16\times15\times14}{5\times4\times3\times2\times1}=8568\)
\(\binom{14}{5}=\frac{14!}{5!(14 - 5)!}=\frac{14\times13\times12\times11\times10}{5\times4\times3\times2\times1}=2002\)

Step 2: Calculate at least one horror

The number of schedules with at least one horror movie is \(\binom{18}{5}-\binom{14}{5}\)
\(8568-2002 = 6566\)

Step 1: Case 1: 4 disaster and 1 horror

We need to choose 4 disaster movies from 14 and 1 horror movie from 4. Using the combination formula \(\binom{n}{k}=\frac{n!}{k!(n - k)!}\), the number of ways for this case is \(\binom{14}{4}\times\binom{4}{1}\)
\(\binom{14}{4}=\frac{14!}{4!(14 - 4)!}=\frac{14\times13\times12\times11}{4\times3\times2\times1}=1001\)
\(\binom{4}{1}=\frac{4!}{1!(4 - 1)!}=\frac{4}{1}=4\)
So, \(\binom{14}{4}\times\binom{4}{1}=1001\times4 = 4004\)

Step 2: Case 2: 5 disaster and 0 horror

The number of ways to choose 5 disaster movies from 14 is \(\binom{14}{5}\)
\(\binom{14}{5}=\frac{14!}{5!(14 - 5)!}=\frac{14\times13\times12\times11\times10}{5\times4\times3\times2\times1}=2002\)

Step 3: Total for at least 4 disaster

We add the number of ways from both cases: \(4004 + 2002=6006\)

Step 1: Choose the two movies

We know that we must include "Airport Disaster" and "Halloween Horror". So we need to choose \(5-2 = 3\) more movies from the remaining \(14 + 4-2=16\) movies.
Using the combination formula \(\binom{n}{k}=\frac{n!}{k!(n - k)!}\), where \(n = 16\) and \(k=3\)
\(\binom{16}{3}=\frac{16!}{3!(16 - 3)!}=\frac{16\times15\times14}{3\times2\times1}=560\)

Answer:

\(6566\)

Part b)