<PRE><B>Question 370920</B>
&lt;br&gt;&lt;font face=&quot;Tahoma&quot;&gt;With this problem, you just need to break it down a bit. &lt;br&gt;

She wants to bring at least one book, but not all five.&lt;br&gt;

This means she wants to bring 1 book, 2 books, 3 books, or 4 books.&lt;br&gt;

All you need to do is set up a different combination for all 4 scenarios described above, and then add them together.&lt;br&gt;

This will give you the total number of subsets with the given constraints:&lt;br&gt;

5C1=5!/(1!*4!)=5&lt;br&gt;

5C2=5!/(2!*3!)=10&lt;br&gt;

5C3=5!/(3!*2!)=10&lt;br&gt;

5C4=5!/(4!*1!)=5&lt;br&gt;

Thus there are 5+10+10+5 different subsets of books she could bring.&lt;br&gt;

Answer: There are 30 different subsets!&lt;br&gt;

I hope this helps!&lt;br&gt;</PRE>