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Puzzle – terms in the expansion

November 15, 2007 am30 9:10 am

ExpandFairly basic, but this is what I do…

How many terms are there in (x+y)^5 ? (x+y)^n ?

How many terms in (x+y+z)^2? (x+y+z)^3? (x+y+z)^5? (x+y+z)^n?

Can you generalize to more terms? (check yourself with (v+w+x+y+z)^7 answer is 210 terms.)

from → Math, Math Education, mathematics, Puzzles
3 Comments leave one →
  1. Clueless permalink
    November 15, 2007 pm30 9:49 pm 9:49 pm

    For (x1+x2+…xk)^n, I get C(n+k-1,k-1) terms.

    This does not agree with your answer for the special case, however. So one of us has made a mistake.

    Additionally, this appears to be a special case of the Bose-Einstein statistics problem of counting the number of ways in which n identical balls can be put in k bins, for which the answer above is correct.

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  2. jd2718 permalink*
    November 16, 2007 am30 7:20 am 7:20 am

    Arrgh, you caught me counting too fast. I meant (v+w+x+y+z)^6

    I think there was an eleven in my mental math (6+5-1), which I dutifully remembered and wrote down, which was wrong.

    Thank you.

    And yes, in k distinct bins.

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  3. jd2718 permalink*
    November 16, 2007 pm30 4:43 pm 4:43 pm

    Here’s the proof I use (generalizing from an example)

    Consider (v+w+x+y+z)^6 The five exponents will add up to 6, so we can consider the equivalent problem a+b+c+d+e=6, where a,b,c,d,e are 0 or positive integers.

    Represent our exponents as stars: ******
    Use vertical lines to split them into groups: ***|*||**|
    This arrangement corresponds to 3,1,0,2,0 or v^3wy^2.
    Notice that lines without stars in between means that that exponent is 0 (as does a line at the beginning or the end).

    All we have to do now is find out how many ways we can rearrange these symbols: C(10,4). Hmm. Where’s the 10 come from? We have 6 exponents, and 6 stars. We have 5 terms, and 4 lines. So exponents + terms – 1 would seem to do it. And the 4? Thats the number of terms – 1.

    So for (x_1+x_2+...+x_k)^n we will have n stars, k – 1 vertical lines, and our answer is Clueless’: C(n+k-1,k-1).

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