Show that p choose n=0 when n>p

  • Thread starter Potatochip911
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In summary: This proves the first statement. In summary, the conversation discusses how to prove that ##\binom{p}{n}=0## when ##n>p##, using the familiar binomial theorem. It is shown that for n≥(p+1), the product of the factors in the binomial coefficient is 0, making the whole coefficient equal to 0. This also leads to the understanding that when expanding ##(1+x)^p##, there will always be ##p+1## terms, with n ranging from 0 to p.
  • #1

Homework Statement


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.

Homework Equations


3. The Attempt at a Solution [/B]
$$\binom{p}{n}=\frac{p(p-1)(p-2)\cdots(p-n+2)(p-n+1)}{n!} $$ 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?
 
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  • #2
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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What does "p choose n" mean?

"P choose n" is a mathematical notation that represents the number of ways to choose a group of n objects from a larger group of p objects, without considering the order in which the objects are chosen. It is also known as a combination.

Why does p choose n equal 0 when n is greater than p?

This is because it is mathematically impossible to choose a group of n objects from a smaller group of p objects when n is larger than p. In other words, there are no combinations possible when n is greater than p, resulting in a value of 0 for "p choose n".

Can p choose n be negative?

No, "p choose n" is always a positive integer. It represents the number of ways to choose a group of n objects from a larger group of p objects, and it is not possible to have a negative number of combinations.

What happens when n and p are equal?

In this case, "p choose n" equals 1. This is because there is only one way to choose a group of n objects from a group of p objects when n and p are equal, and that is by choosing all p objects.

Can "p choose n" be calculated using a formula?

Yes, the formula for "p choose n" is nCr = p! / (n!(p-n)!), where p! represents the factorial of p. This formula can be used to calculate the number of combinations when n is less than or equal to p.

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