<p>Is this even possible. Barron’s doesn’t explain how to do it. For example, it says find the first 3 terms of (a+b)^(-3). This is on page 89 by the way. I get how to find the nth term for positive integer exponents. Willl we get non-positive integers or non-integer positive numbers as exponents for these types of problems on the Math II?</p>
<p>Yes, combinations with negative numbers (and fractions!) do exist.</p>
<p>For example, (n choose -3) = (n)(n-1)(n-2) / [ (-3)(-2)(-1) ]</p>
<p>It is rather ridiculous that Barron’s has this in their prep book. I have never seen it in a school textbook and it will definitely not appear on the test. (It does appear in Art of Problem Solving textbook, though, so it may appear on AMC/AIME. 
)</p>
<p>Wouldn’t that be (-3)(-4)(-5) since you’re subtracting 1 form n each time.</p>
<p>oops, my bad.</p>
<p>Actually, when doing (n choose k), k MUST be positive.</p>
<p>It is n that could be fractional or negative.</p>