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Show that p choose n=0 when n>p

  1. May 24, 2015 #1
    1. The problem statement, all variables and given/known data
    Show that if p is a positive integer, then ##\binom{p}{n}=0## when ##n>p##, so ##(1+x)^p=\sum \binom{p}{n} x^n ##, is just a sum of ##p+1## terms, from ##n=0## to ##n=p##. For example ##(1+x)^2## has 3 terms, ##(1+x)^3## has 4 terms, etc. This is just the familiar binomial theorem.

    2. Relevant equations
    3. The attempt at a solution

    $$\binom{p}{n}=\frac{p(p-1)(p-2)\cdots(p-n+2)(p-n+1)}{n!} $$ Since n>p I figured I would test with (p+1) to see what happens. $$\binom{p}{p+1}=\frac{p(p-1)(p-2)\cdots(p-(p+1)+2)(p-(p+1)+1)}{(p+1)!}$$ So the last term clearly goes to zero. Is rewriting ##\binom{p}{n}=0## as $$\frac{(p-n+1)(p-n+2)(p-n+3)(p-n+4)\cdots p}{n!} $$ sufficient for proving this? I am also quite lost as to what they're implying with $$(1+x)^p=\sum \binom{p}{n} x^n $$ I understand that there is always ##(p+1)## terms after expanding a binomial raised to a power but by the sum are they referring to if the binomial was rewritten as a Maclaurin series?
     
  2. jcsd
  3. May 25, 2015 #2

    Svein

    User Avatar
    Science Advisor

    You are so close! You have already [itex]
    \begin{pmatrix}
    p\\
    n\\
    \end{pmatrix}=\frac{p(p-1)(p-2)\dotso (p-n+1)}{n!}
    [/itex]. Therefore [itex]
    \begin{pmatrix}
    p\\
    p\\
    \end{pmatrix}=\frac{p(p-1)(p-2)\dotso (p-p+1)}{p!}=1
    [/itex] and [itex]
    \begin{pmatrix}
    p\\
    p+1\\
    \end{pmatrix}=\frac{p(p-1)(p-2)\dotso (p-(p+1)+1)}{(p+1)!}=0
    [/itex]. So, for n≥(p+1), you have one factor = 0, therefore the product is 0.
     
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