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$$\frac{(n+1)!}{k! \cdot (n-k+1)!}$$

I don't have any clue. Can $(n-k+1)!$ be simplified into $(n+1)!(k)!$

I am not too familiar with this symbol: !.

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    $\begingroup$ Relevant reading: Factorial and Binomial_coefficient $\endgroup$ – user2468 Nov 1 '12 at 16:33
  • $\begingroup$ The symbol $!$ refers to factorial. $\endgroup$ – Sasha Nov 1 '12 at 16:34
  • $\begingroup$ i get n!(n+1)/k! (n-k)! (n-k+1) how do I get rid of the k in the denominator? $\endgroup$ – Martin Desjardins Nov 1 '12 at 16:47
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This is the binomial coefficient $\binom{n+1}{k}$.

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  • $\begingroup$ how did you get to that? $\endgroup$ – Martin Desjardins Nov 1 '12 at 16:51
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    $\begingroup$ The formula for $\binom{n}{k}$ is $\frac{n!}{k!(n-k)!}$. If you substitute $n+1$ for $n$, you get the above. $\endgroup$ – ngn Nov 1 '12 at 16:53
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The symbol $!$ is the factorial, it is defined as $$n! = n\cdot(n-1)\cdot(n-2)\cdots 2\cdot 1 = \prod_{j=1}^n j.$$ The expression $\frac{n!}{k!\cdot(n-k)!}=\binom nk$ is also known as the binomial coefficient. There is a number of interesting properties of this quantity. Most intuitively, it is the number of ways to pick $k$ elements out of a set that contains $n$ elements.

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A concrete example might be helpful. Let $n = 4$ and $k = 3$. The expression can be expanded to $$ \frac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 }{(3 \cdot 2 \cdot 1)(2 \cdot 1)}. $$

Notice both the expressions in the denominator are shorter factorials, so you can quickly cancel the "tail" of the factorial in the numerator as $$ \frac{5 \cdot 4}{2 \cdot 1} $$ or $$ \frac{5 \cdot 4 \cdot 3}{3 \cdot 2 \cdot 1}. $$ (Of course, more cancellation can be done with specific numbers, but I'm trying to mimic the general case where $n$ and $k$ are not specified.)

Back to general $n$ and $k$, we can simplify your original expression as either $$ \frac{(n+1) \cdot n \cdots (k+1)}{(n+1-k)!} $$ or $$ \frac{(n+1) \cdot n \cdots (n+2-k)}{k!}. $$ In the former, I cancelled the $k!$ "tail". In the latter, I cancelled the $(n+1-k)!$ "tail". Without more information about $n$ and $k$, this is the most cancellation one can hope for.

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