Friday, October 2, 2009

(x+y+z)^100

How many terms does (x+y+z)^100 have when expanded? (After combining like terms.)
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1 comment:

  1. DavidOctober 3, 2009 at 3:10 PM

    The answer is 5151.

    Each term has the form A x^i y^j z^k where i, j, k are non-negative integers and i + j + k = 100.

    Note that i can take any value between 0 and 100.

    If i is given, then j can take any value between 0 and 100-i, and k is determined by i and j. So for each i, there are 101-i possibilities.

    Thus the total number of terms is the sum of (101-i) from i = 0 to 100, which is
    101 + 100 + 99 + ... + 1 + 0 = 101*102/2 = 5151.

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