Question 370920: Hi. I am having trouble figuring out this problem. Please help me understand it.
Martina brought five books to bring on vacation. However she decided that her bag was too heavy and that she could not bring all of them. How many possible subsets if she brings at least one, but not all five?
Answer by neatmath(302)   (Show Source):
You can put this solution on YOUR website!
With this problem, you just need to break it down a bit.
She wants to bring at least one book, but not all five.
This means she wants to bring 1 book, 2 books, 3 books, or 4 books.
All you need to do is set up a different combination for all 4 scenarios described above, and then add them together.
This will give you the total number of subsets with the given constraints:
5C1=5!/(1!*4!)=5
5C2=5!/(2!*3!)=10
5C3=5!/(3!*2!)=10
5C4=5!/(4!*1!)=5
Thus there are 5+10+10+5 different subsets of books she could bring.
Answer: There are 30 different subsets!
I hope this helps!
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