Question Re: Permutation with Exceptions

[TunaInTheBrine]

I'm having a really hard time wrapping my head around this one...

QUESTION:

There are 8 contestants who are going to be awarded 4 travel package prizes: Bermuda, Paris, Honolulu, and another Honolulu (i.e. two of the four packages are the same). In how many unique ways can these 8 contestants be awarded these travel packages, taking into consideration that two of them are the same?

My Answer (I don't know if it's right):

8! / 4! 2! = 840

 

[TunaInTheBrine]

Combinatorics Question with Exceptions

Q: There are 7 individual contestants and 4 travel award packages - Jamaica, Bahama, and two in Honolulu (note* there are only 3 unique travel award packages). In how many unique ways can these award packages be distributed?

Okay, so this is essentially a question about combinations with exceptions. If someone asked me how many unique ways can you arrange the letters in the word 'CHILL', I would say it is 5!/2! since there are 5 letters and 2 exceptions. But in this problem, I am confused by the addition of the 7 contestants. Applying the previous example of the word 'CHILL' to the awards package, I would say there are 4!/2! = 12 unique arrangements. So then do I just multiply 12 x 7?