Combinations Problem: Solving 12 Friends, 6 Movies Puzzle

[TbbZz]

Homework Statement

A group of 12 friends goes to a cinema complex that is showing 6 different movies. If the group splits up into subgroups based on movie preferences, how many subgroup combinations are possible?

Homework Equations

nCr
nPr

(we can use calculators)

The Attempt at a Solution

I tried all sorts of ways to solve this problem using nCr and nPr but I couldn't seem to figure it out. I also tried using the equations that involve repetitions, but they didn't seem to help. I also tried simply punching in (relevant) numbers for twenty minutes to help me figure out what functions to use.

The correct answer is 6188, and I was unable to get this answer. I am looking for guidance on answering this problem. Thanks in advance for the help.

 

[sutupidmath]

is a person allowed to whatch more than one movie?

 

[TbbZz]

No, they only watch one movie.

 

[Shooting Star]

A group of 12 friends goes to a cinema complex that is showing 6 different movies. If the group splits up into subgroups based on movie preferences, how many subgroup combinations are possible?

If the people are considered identical, then this reduces to the problem of distributing 12 identical objects in six different boxes in all possible ways, where a box may contain 0 to 12 objects, in which case the answer is (6+12-1)C12 = 17C5 =6188.

However, as the problem is given, this answer is incorrect. The correct treatment is a bit more complicated.

 

[TbbZz]

If the people are considered identical, then this reduces to the problem of distributing 12 identical objects in six different boxes in all possible ways, where a box may contain 0 to 12 objects, in which case the answer is (6+12-1)C12 = 17C5 =6188.

However, as the problem is given, this answer is incorrect. The correct treatment is a bit more complicated.

The people are considered identical.

While I understand that 17C12 equals 6188, would you mind clarifying where the numbers came from?

for the notation: nCr , I understand that the "r" equals 12; however, I'm not sure how you determined the (6+12-1) part.

Thanks for the help.

 

[Shooting Star]

The people are considered identical.
for the notation: nCr , I understand that the "r" equals 12; however, I'm not sure how you determined the (6+12-1) part.

This is a standard result, the proof of which is found in any book of combinatorics dealing with problems of moderate level of difficulty. If 'r' identical objects are distributed in 'n' different boxes, then the number of ways of doing it is

^{n+r-1}C_{n} = ^{n+r-1}C_{r-1}.

I am referring you http://books.google.co.in/books?id=...=EEa64bdr1puHFWd1bSNLSq2DDVk&hl=en#PPA102,M1", a random source I found in the net. Read pages 102-103. If you still have doubts, you can ask me.

 

[HallsofIvy]

If I understand the problem correctly, that the people can divide into any number of subgroups with each subgroup watching one of the movies, then the problem is quite a bit more complicated.

For example, if they remain in a single group (the only subset is the entire set) then they could watch anyone of the six movies: that's 6 possibilities there. Or each could go to a different movie ("singleton" subgroups) and there are 6!ways to do that.

The first question then is: how many ways can you partition of set of 6?

 

[Shooting Star]

The first question then is: how many ways can you partition of set of 6?

Hi HallsofIvy,

I had expressed my misgivings about the method of solving this (post #4), but the OP says that the people are considered to be identical (post #5). In that case, there is no need to consider the number of ways a set of 6 can be partitioned, and the solution is straight forward (post #4).